GPS Pre-Calculus
Frameworks
Teacher’s Edition
Unit 2
Sequences and Series
1st Edition
April, 2011
Georgia Department of Education
Georgia Department of Education
GPS Pre-Calculus
Unit 2
1st Teacher’s Edition
Table of Contents
INTRODUCTION: ..................................................................................................3
Notes on Renaissance Festival Learning Task ......................................................7
Notes on Fascinating Fractals Learning Task.....................................................21
Notes on Diving into Diversions Learning Task .................................................39
Notes on Interior Design Interview Culminating Task ......................................48
Student Version: Interior Design Interview ........................................................54
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Unit 2
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GPS Pre-Calculus – Unit 2
Sequences and Series
Teacher’s Edition
INTRODUCTION:
In earlier grades, students learned about arithmetic and geometric sequences and their
relationships to linear and exponential functions, respectively. This unit builds on students’
understandings of those sequences and extends students’ knowledge to include arithmetic and
geometric series, both finite and infinite. Summation notation and properties of sums are also
introduced. Additionally, students will examine other types of sequences and, if appropriate,
proof by induction. They will use their knowledge of the characteristics of the types of sequences
and the corresponding functions to compare scenarios involving different sequences.
ENDURING UNDERSTANDINGS:

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All arithmetic and geometric sequences can be expressed recursively and explicitly.
Some other sequences also can be expressed in both ways but others cannot.
Arithmetic sequences are identifiable by a common difference and can be modeled by
linear functions. Infinite arithmetic series always diverge.
Geometric sequences are identifiable by a common ratio and can be modeled by
exponential functions. Infinite geometric series diverge if r  1 and converge is r  1 .
The sums of finite arithmetic and geometric series can be computed with easily derivable
formulas.
Identifiable sequences and series are found in many naturally occurring objects.
Repeating decimals can be expressed as fractions by summing appropriate infinite
geometric series.
KEY STANDARDS ADDRESSED:
MM4A9. Students will use sequences and series
a. Use and find recursive and explicit formulae for the terms of sequences.
b. Recognize and use simple arithmetic and geometric sequences.
c. Find and apply the sums of finite and, where appropriate, infinite arithmetic
and geometric series.
d. Use summation notation to explore finite series.
RELATED STANDARDS ADDRESSED:
MM4A1. Students will explore rational function.
a. Investigate and explore characteristics of rational functions, including domain, range,
zeros, points of discontinuity, intervals of increase and decrease, rates of change, local
and absolute extrema, symmetry, asymptotes, and end behavior.
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MM4A4. Students will investigate functions.
a. Compare and contrast properties of functions within and across the following types:
linear, quadratic, polynomial, power, rational, exponential, logarithmic, trigonometric,
and piecewise.
b. Investigate transformations of functions.
MM4P1. Students will solve problems (using appropriate technology).
a. Build new mathematical knowledge through problem solving.
b. Solve problems that arise in mathematics and in other contexts.
c. Apply and adapt a variety of appropriate strategies to solve problems.
d. Monitor and reflect on the process of mathematical problem solving.
MM4P2. Students will reason and evaluate mathematical arguments.
a. Recognize reasoning and proof as fundamental aspects of mathematics.
b. Make and investigate mathematical conjectures.
c. Develop and evaluate mathematical arguments and proofs.
d. Select and use various types of reasoning and methods of proof.
MM4P3. Students will communicate mathematically.
a. Organize and consolidate their mathematical thinking through communication.
b. Communicate their mathematical thinking coherently and clearly to peers, teachers, and
others.
c. Analyze and evaluate the mathematical thinking and strategies of others.
d. Use the language of mathematics to express mathematical ideas precisely.
MM4P4. Students will make connections among mathematical ideas and to other
disciplines.
a. Recognize and use connections among mathematical ideas.
b. Understand how mathematical ideas interconnect and build on one another to produce a
coherent whole.
c. Recognize and apply mathematics in contexts outside of mathematics.
MM4P5. Students will represent mathematics in multiple ways.
a. Create and use representations to organize, record, and communicate mathematical ideas.
b. Select, apply, and translate among mathematical representations to solve problems.
c. Use representations to model and interpret physical, social, and mathematical
phenomena.
UNIT OVERVIEW:
The launching activity begins by revisiting ideas of arithmetic sequences studied in eighth and
ninth grades. Definitions, as well as the explicit and recursive forms of arithmetic sequences are
reviewed. The task then introduces summations, including notation and operations with
summations, and summing arithmetic series.
The second set of tasks reviews geometric sequences and investigates sums, including infinite
and finite geometric series, in the context of exploring fractals. It is assumed that students have
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some level of familiarity with geometric sequences and the relationship between geometric
sequences and exponential functions.
The third group addresses some common sequences and series, including the Fibonacci
sequence, sequences with factorials, and repeating decimals.
The culminating task is set in the context of applying for a job at an interior design agency. In
each task, students will need to determine which type of sequence is called for, justify their
choice, and occasionally prove they are correct. Students will complete the handshake problem,
the salary/retirement plan problem, and some open-ended design problems that require the use of
various sequences and series.
VOCABULARY AND FORMULAS
Arithmetic sequence: A sequence of terms a1 , a2 , a3 ,... with d  an  an1 . The explicit formula
is given by an  a1   n  1 d and the recursive form is a1 = value of the first term and
an  an 1  d .
Arithmetic series: The sum of a set of terms in arithmetic progression a1  a2  a3  ... with
d  an  an1 .
Common difference: In an arithmetic sequence or series, the difference between two
consecutive terms is d, d  an  an1 .
Common ratio: In a geometric sequence or series, the ratio between two consecutive terms is r,
a
r n .
an 1
Explicit formula: A formula for a sequence that gives a direct method for determining the nth
term of the sequence. It presents the relationship between two quantities, i.e. the term number
and the value of the term.
Factorial: If n is a positive integer, the notation n! (read “n factorial”) is the product of all
positive integers from n down through 1; that is, n!  n  n  1 n  2 ... 3 21 . Note that 0!, by
definition, is 1; i.e. 0!  1.
Finite series: A series consisting of a finite, or limited, number of terms.
Infinite series: A series consisting of an infinite number of terms.
Geometric sequence: A sequence of terms a1 , a2 , a3 ,... with r 
an
. The explicit formula is
an 1
given by an  a1r n 1 and the recursive form is a1  value of the first term and an   an1  r .
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Geometric series: The sum of a set of terms in geometric progression a1  a2  a3  ... with
a
r n .
an 1
Limit of a sequence: The long-run value that the terms of a convergent sequence approach.
Partial sum: The sum of a finite number of terms of an infinite series.
Recursive formula: Formula for determining the terms of a sequence. In this type of formula,
each term is dependent on the term or terms immediately before the term of interest. The
recursive formula must specific at least one term preceding the general term.
Sequence: A sequence is an ordered list of numbers.
n
Summation or sigma notation:
 a , where i is the index of summation, n is the upper limit of
i 1
i
summation, and 1 is the lower limit of summation. This expression gives the partial sum, the sum
n
of the first n terms of a sequence. More generally, we can write
 a , where k is the starting
ik
i
value.
Sum of a finite arithmetic series: The sum, Sn, of the first n terms of an arithmetic sequence is
n  a1  an 
given by Sn 
, where a1 = value of the first term and an = value of the last term in the
2
sequence.
Sum of a finite geometric series: The sum, Sn, of the first n terms of a geometric sequence is
n
a1  an r a1 1  r

given by Sn 
, where a1 is the first term and r is the common ratio (r  1).
1 r
1 r


Sum of an infinite geometric series: The general formula for the sum S of an infinite geometric
a
series a1  a2  a3  ... with common ratio r where r  1 is S  1 . If an infinite geometric
1 r
series has a sum, i.e. if r  1 , then the series is called a convergent geometric series. All other
geometric (and arithmetic) series are divergent.
Term of a sequence: Each number in a sequence is a term of the sequence. The first term is
generally noted as a1, the second as a2, …, the nth term is noted as an. an is also referred to as the
general term of a sequence.
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Notes on Renaissance Festival Learning Task
This task launches the unit by reviewing students’ understanding of arithmetic sequences and
building on that knowledge to examine arithmetic series. Students will review arithmetic
sequences by investigating the radii and areas on an archery target and in determining how
many jars are needed in the Rock Throwing contest. Recursive forms will be reviewed, as well
as the explicit forms, both the connection to the slope-intercept form of a line and the common
difference form. This task will introduce summing series and summation notation. Properties
of sums, including the impossibility of summing an infinite arithmetic series, will also be
addressed.
Supplies Needed:
 Graph paper
 Calculators
RENAISSANCE FESTIVAL LEARNING TASK
As part of a class project on the Renaissance, your class decided to plan a renaissance festival for
the community. Specifically, you are a member of different groups in charge of planning two of
the contests. You must help plan the archery and rock throwing contests. The following activities
will guide you through the planning process.
Group One: Archery Contest1
Before planning the archery contest, your group decided to investigate the characteristics of the
target. The target being used has a center, or bull’s-eye, with a radius of 4 cm, and nine rings that
are each 4 cm wide.
1. The Target
a. Sketch a picture of the center and first 3 rings of the target.
b. Write a sequence that gives the radius of each of the concentric circles that comprise the
entire target.
4, 8, 12, 16, 20, 24, 28, 32, 36, 40 (All measures are in terms of centimeters.)
1
Elements of these problems were adapted from Integrated Mathematics 3 by McDougal-Littell,
2002.)
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c. Write a recursive formula and an explicit formula for the terms of this sequence.
a1  4
and an  4n .

an  an1  4
d. What would be the radius of the target if it had 25 rings? Show how you completed this
problem using the explicit formula.
a25  4(25)  100cm
e. In the past, you have studied both arithmetic and geometric sequences. What is the
difference between these two types of sequences? Is the sequence in (b) arithmetic,
geometric, or neither? Explain.
Arithmetic sequences – the terms are found by adding or subtract a common number to
each subsequent term. These are modeled by linear functions. Geometric sequences –
the terms are found by multiplying a common ratio to each subsequent term. These are
modeled with exponential functions. The sequence in this problem is arithmetic; we
can add 4 cm to each term to get the next term.
One version of the explicit formula uses the first term, the common difference, and the number
of terms in the sequence. For example, if we have the arithmetic sequence 2, 5, 8, 11, 14, …, we
see that the common difference is 3. If we want to know the value of the 20th term, or a20, we
could think of starting with a1 = 2 and adding the difference, d = 3 a certain number of times.
How many times would we need to add the common difference to get to the 20th term? _____
Because multiplication is repeated addition, instead of adding 3 that number of times, we could
multiply the common difference, 3, by the number of times we would need to add it to 2.
This gives us the following explicit formula for an arithmetic sequence: an  a1   n  1 d .
f. Write this version of the explicit formula for the sequence in this problem. Show how this
version is equivalent to the version above.
an  4  4(n  1)  4  4n  4  4n
g. Can you come up with a reason for which you would want to add up the radii of the
concentric circles that make up the target (for the purpose of the contest)? Explain.
Students may come up with a variety of answers. Generally, there is not a good reason.
Finding the radius of the last circle incorporates the other radii.
h. Plot the sequence from this problem on a coordinate grid. What should you use for the
independent variable? For the dependent variable? What type of graph is this? How does
the an equation of the recursive formula relate to the graph? How does the parameter d in
the explicit form relate to the graph?
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The independent variable is the term number; the dependent is the value of the term.
This is a linear graph. The an equation of the recursive form tells you to go up 4 and
over 1. The d is the slope of the line.
Term Value (Radius)
Target Radii
40
35
30
25
20
15
10
5
0
0
2
4
6
8
10
Term Number
i. Describe (using y-intercept and slope), but do not graph, the plots of the arithmetic
sequences defined explicitly or recursively as follows:
4
1. an  3   n  1
3. an  4.5  3.2  n  1
3
a1  10
a1  2


2. 
4. 
2
1
an  an 1  2
an  an 1  5
1. The y-intercept of the graph is 3 – 4/3 = -1/3. The slope is 4/3.
2. The y-intercept is -5/2. The slope is ½.
3. The y-intercept is 4.5 + 3.2 = 7.7. The slope is -3.2.
4. The y-intercept is 10 2/5. The slope is -2/5.
2. The Area of the Target: To decide on prizes for the archery contest, your group decided to use
the areas of the center and rings. You decided that rings with smaller areas should be worth more
points. But how much more? Complete the following investigation to help you decide.
a. Find the sequence of the areas of the rings, including the center. (Be careful.)
16, 48, 80, 112, 144, 176, 208, 240, 272, 304
(All areas are in square cm.)
b. Write a recursive formula and an explicit formula for this sequence.
a1  16

an  an1  32
an  16   n  1 32
or an  32 n  16 or an  16  2n  1
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c. If the target was larger, what would be the area of the 25th ring?
a25  16  24(32 )  784 cm 2
d. Find the total area of the bull’s eye by adding up the areas in the sequence.
1600 cm2
e. Consider the following sum: Sn  a1  a2  a3  ...  an1  an2  an . Explain why that
equation is equivalent to Sn  a1   a1  d    a1  2d   ...   an  2d    an  d   an .
Rewrite this latter equation and then write it out backwards. Add the two resulting
equations. Use this to finish deriving the formula for the sum of the terms in an arithmetic
sequence. Try it out on a few different short sequences.
The equations are the same because the nth term can be found by adding n-1
differences. Alternately, the appropriate number of differences can be subtracted from
the last term.
Sn  a1   a1  d    a1  2d   ...   an  2d    an  d   an
Sn  an   an  d    an  2d   ...   a1  2d    a1  d   a1
2Sn   a1  an  n
Sn 
n  a1  an 
2
f. Use the formula for the sum of a finite arithmetic sequence in part (e) to verify the sum of
the areas in the target from part (d).
Sn 
n  a1  an 
2

10 16  304 
2
 1600 cm2
g. Sometimes, we do not have all the terms of the sequence but we still want to find a
specific sum. For example, we might want to find the sum of the first 15 multiples of 4.
Write an explicit formula that would represent this sequence. Is this an arithmetic
sequence? If so, how could we use what we know about arithmetic sequences and the
sum formula in (e) to find this sum? Find the sum.
an = 4n. This is arithmetic. We use the explicit form to find the 15th term. a15 = 60; S15 =
(15/2)(4 + 60) = (7.5)(64) = 480.
Comment:
Teachers can, at some point, show students how to use the TI calculator to sum
sequences. Detailed instructions are available in most textbooks and should be
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available in a TI handbook. The entry screen for this last sum would be
sum(seq(4x,x,1,15,1)). This tells the calculator that you want to find the sum of the
sequence 4x, where x is the independent variable. You are finding the sum of the terms
from a1 to a15, including each term between. If you only wanted to include the odd
terms, i.e. a1, a3, etc., the last number would be 2 because you are counting up by 2s.
h. What happens to the sum of the arithmetic series we’ve been looking at as the number of
terms we sum gets larger? How could you find the sum of the first 200 multiples of 4?
How could you find the sum of all the multiples of 4? Explain using a graph and using
mathematical reasoning.
The sum gets larger and larger. To find the sum of the first 200 multiples of 4, we need
to first find the 200th multiple, 4(200) = 800. Then we use the formula (200/2)(4+800) =
100(804) = 80,400. We could not find the sum of all the multiples; it keeps getting
larger. If you look at a graph of an arithmetic sequence, the line increases without
bound.
i. Let’s practice a few arithmetic sum problems.
1. Find the sum of the first 50 terms of 15, 9, 3, -3, …
2. Find the sum of the first 100 natural numbers
3. Find the sum of the first 75 positive even numbers
4. Come up with your own arithmetic sequence and challenge a classmate to find the
sum.
1. a50 = 15 + (49)(-6) = -279; S50 = 25(15 + -279) = -6600
2. a100 = 100; S100 = 50(1 + 100) = 5050
3. a75 = 2 + (74)2 = 150; S75 = (75/2)(2 + 150) = 5700
4. Answers will vary.
j. Summarize what you learned / reviewed about arithmetic sequences and series during this
task.
Answers will vary but should include the explicit formula for the terms of an arithmetic
sequence, the formula for the sum of an arithmetic series, that arithmetic sequences can
be modeled by linear functions (although they are discrete, not continuous), and that we
cannot find the sum of an infinite arithmetic series.
3. Point Values: Assume that each participant’s arrow hits the surface of the target.
a. Determine the probability of hitting each ring and the bull’s-eye.
Target Piece
Area of Piece
Probability of
(in cm2)
Hitting this Area
Bull’s Eye
16/1600 = .01
16
Ring 1
48/1600 = .03
48
Ring 2
80/1600 = .05
80
Ring 3
112/1600 = .07
112
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Ring 4
Ring 5
Ring 6
Ring 7
Ring 8
Ring 9
144
176
208
240
272
304
1st Teacher’s Edition
144/1600 = .09
176/1600 = .11
208/1600 = .13
240/1600 = .15
272/1600 = .17
304/1600 = .19
b. Assign point values for hitting each part of the target, justifying the amounts based on the
probabilities just determined.
These answers will vary. Prize values should decrease as one moves out from the bull’s
eye. One arrangement could be, from the bull’s eye to the outer ring, 50, 45, 40, 35, etc.
c. Use your answer to (b) to determine the expected number of points one would receive after
shooting a single arrow.
Again, answers will vary. If the proposed solution to (b) is used, the expected value
would be 50(.01) + 45(.03) + 40(.05) + … = 19.25 points.
d. Using your answers to part (c), determine how much you should charge for participating in
the contest OR for what point values participants would win a prize. Justify your
decisions.
Answers will vary.
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Group Two: Rock Throwing Contest2
Comment:
The third part of this problem moves beyond arithmetic sequences and series. This part could
be used as an extension or for differentiating for gifted students. The fourth part includes
practice with summation notation.
For the rock throwing contest, your group decided to provide three different arrangements of
cans for participants to knock down.
1. For the first arrangement, the tin cans were set up in a triangular pattern,
only one can deep. (See picture.)
a. If the top row is considered to be row 1, how many cans would be on
row 10?
Approaches may vary. Students should realize that the nth row has n
cans. So row 10 would have 10 cans.
b. Is this an arithmetic or a geometric sequence (or neither)? Write explicit
and recursive formulas for the sequence that describes the number of
cans in the nth row of this arrangement.
This is an arithmetic sequence. The number of cans increases by 1 each time you move
to another row.
a1  1
an  n

an  an 1  1
c. It is important to have enough cans to use in the contest, so your group needs to determine
how many cans are needed to make this arrangement. Make a table of the number of rows
included and the total number of cans.
Rows Included
1
2
3
4
5
6
7
8
Total Cans
1
3
6
10
15
21
28
36
2
Adapted from Manouchehri, A. (2007). Inquiry-discourse mathematics instruction.
Mathematics Teacher, 101, 290–300.
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Comment:
An extension here could include looking at finite differences, noting that the total cans
could be modeled with a quadratic function, given that the second differences are
constant.
d. One of your group members decides that it would be fun to have a “mega-pyramid” 20
rows high. You need to determine how many cans would be needed for this pyramid, but
you don’t want to add all the numbers together. One way to find the sum is to use the
summation formula you found in the Archery Contest. How do you find the sum in an
arithmetic sequence? ____________ Find the sum of a pyramid arrangement 20 rows
high using this formula.
Sn 
n  a1  an 
2
. S20 = 10(1 + 20) = 210 cans
n
We can also write this problem using summation notation:
 a , where i is the index of
i 1
i
summation, n is the upper limit of summation, and 1 is the lower limit of summation. We can
20
think of ai as the explicit formula for the sequence. In this pyramid problem, we have
i
i 1
because we are summing the numbers from 1 to 20. We also know what this sum is equal to.
20
n
20
i   a1  an   1  20  . What if we did not know the value of n, the upper limit but we

2
2
i 1
did know that the first number is 1 and that we were counting up by 1s? We would then have
n
n  n  1
n
. This is a very common, important formula in sequences. We will use
i  1  n  

2
2
i 1
it again later.
e. Propose and justify a specific number of cans that could be used in this triangular
arrangement. Remember, it must be realistic for your fellow students to stand or sit and
throw a rock to knock down the cans. It must also be reasonable that the cans could be set
back up rather quickly. Consider restricting yourself to less than 50 cans for each
pyramid. Describe the set-up and exactly how many cans you need.
Answers will vary. Be sure that the solution is reasonable and correctly described and
calculated.
2. For the second arrangement, the group decided to make another triangular
arrangement; however, this time, they decided to make the pyramid 2 or 3
cans deep. (The picture shows the 2-deep arrangement.)
a. This arrangement is quite similar to the first arrangement. Write an
explicit formula for the sequence describing the number of cans in the
nth row if there are 2 cans in the top row, as pictured.
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an = 2n, where n is the row number
b. Determine the number of cans needed for the 20th row.
a20 = 2(20) = 40 cans
c. Similar to above, we need to know how many cans are needed for this arrangement. How
will this sum be related to the sum you found in problem 1?
Because it is the same arrangement twice, the number of cans needed should be twice
the number of cans needed in problem 1.
d. The formula given above in summation notation only applies when we are counting by
ones. What are we counting by to determine the number of cans in each row? What if the
cans were three deep? What would we be counting by? In this latter case, how would the
sum of the cans needed be related to the sum of the cans needed in the arrangement in
problem 1?
In this problem, we are counting by twos. If the cans were three deep, we would need to
count by threes. In this latter case, the sum of the cans needed would be three times the
sum of the cans needed in problem 1.
n
n
i 1
i 1
e. This leads us to an extremely important property of sums:  cai  c  ai , where c is a
constant. What does this property mean? Why is it useful?
This property says that if we are summing a sequence in which the explicit formula
includes multiplying by constant, we can find the sum of the sequence without the
constant and then multiply by the constant at the end. This is useful because we
already have a formula for the sum of the first n counting numbers.
f. Suppose you wanted to make an arrangement that is 8 rows high and 4 cans deep. Use the
property in 2e to help you determine the number of cans you would need for this
arrangement.
8
We would need the answer to the following:
following:
8
8
i 1
i 1
 8 1  8 
  144 .
2


 4i . Using the property, we get the
i 1
 4i  4 i  4 
g. Propose and justify a specific number of cans that could be used in this triangular
arrangement. You may decide how many cans deep (>1) to make the pyramid. Consider
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restricting yourself to less than 50 cans for each pyramid. Describe the set-up and exactly
how many cans you need. Show any calculations.
Answers will vary. Be sure that the solution is reasonable and correctly described and
calculated.
3. For the third arrangement, you had the idea to make the pyramid of
cans resemble a true pyramid. The model you proposed to the group
had 9 cans on bottom, 4 cans on the second row, and 1 can on the
top row.
a. Complete the following table.
Row
Number
of Cans
1
2
3
4
5
6
7
8
9
10
1
4
9
16
25
36
49
64
81
100
Change
from
Previous
Row
1
3
5
7
9
11
13
15
17
19
b. How many cans are needed for the nth row of this arrangement?
The nth row has n2 cans.
c. What do you notice about the numbers in the third column above? Write an equation that
relates column two to column three. Then try to write the equation using summation
notation.
Comment:
Students may need help coming up with the explicit form of an odd number. Prior
discussion about how to express even and odd numbers may be needed, perhaps as the
warm-up for the day.
Solution:
Students should notice a few things about the numbers. First, they differ by 2, so they
can be expressed using a linear function. Second, they consist of the positive odd
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integers (or the odd natural numbers). n2 = 1 + 3 + 5 + … + (2n – 1) and
n
  2i  1  n
2
.
i 1
d. How could you prove the relationship you identified in 3c?
Comment:
This provides students an opportunity to think about proof outside of geometry, to
make conjectures, and to test out their conjectures. The next few questions provide
some possible proof options. Students need opportunities to see proofs that are not twocolumn or paragraph geometry proofs. These examples provide such opportunities.
Solution:
Answers will vary.
e. Let’s look at a couple of ways to prove this relationship. Consider a visual approach to a
proof.3 Explain how you could use this approach to prove the relationship.
One could illustrate that this pattern would continue, resulting in a pattern similar to
the next one. The figures illustrate that the number of dots contained in each n x n
array represent the sum of the first n odd natural numbers. In the general case, the
number of dots in the nth array is n2.
3
A similar approach can be found in the August 2006 issue of the Mathematics Teacher:
Activities for Students: Visualizing Summation Formulas by Gunhan Caglayan.
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f. We know that the sum of the first n natural number is
1  2  3  ...  n 
n  n  1
n  n  1
2
, so
. If we multiply both sides of the equation by 2, we get the sum
2
of the first n EVEN numbers. How can we use this new equation to help us find the sum
of the first n ODD numbers?
Multiplying both sides by 2 yields 2 + 4 + 6 + … + 2(n – 1) + 2n = n(n + 1). If we
subtract 1 from each of these n even numbers, we will obtain the sum of the first n odd
natural numbers.
1 + 3 + 5 + 2(n – 1) – 1 + 2n – 1 = n(n + 1) – 1(n) = n(n + 1) – n = n2 + n – n = n2.
g. Consider another approach. We have S  n   1  3  5  ...   2n 1 . If we reverse the
ordering in this equation, we get S  n    2n  1  ...  5  3  1. What happens if we add
the corresponding terms of these two equations? How will that help us prove the
relationship we found earlier?
We get the following from adding the corresponding terms:
2S(n) = (1 + 2n – 1) + (3 + 2(n-1) – 1) + … + (2(n-1) – 1 + 3) + (2n – 1 + 1)
= 2n + 2n + … + 2n = n(2n) = 2n2
However, we only want S(n), so we can divide both sides by 2 and get the desired
S(n) = n2.
h. In a future task, you will learn another way to prove this relationship. You will also look
at the sum of rows in this can arrangement. Can you conjecture a formula for the sum of
the first n square numbers? Try it out a few times.
Comment:
Students may need to be pushed in the correct direction. One possible avenue is for
them to use finite differences when looking at the sequence of sums. For example, the
sums are 1, 5, 14, 30, …. The third differences are all 2, a constant, indicating that a
cubic equation will be needed to determine the nth sum.
Solution:
n
Answers will vary. The actual sum is
i
i 1
2

n  n  1 2n  1
6
.
4. Throughout this task, you learned a number of facts and properties about summation notation
and sums. You learned what summation notation is and how to compute some sums using the
notation. There is another important property to learn that is helpful in computing sums.
We’ll look at that here, along with practicing using summation notation.
a. Write out the terms of these series.
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4
i.
 5i
i 1
5
ii.
i  6
i 1
5
iii.
i
2
3
i 1
i. 5 + 10 + 15 + 20 = 50
ii. 7 + 8 + 9 + 10 + 11 = 45
iii. 4 + 7 + 12 + 19 + 28 = 70
b. You have already seen one sum property. Here are the important properties you need to
know. Explain why the two new properties make mathematical sense to you.
Properties of sums (c represents a constant)
1.
2.
n
n
i 1
n
i 1
 cai  c ai
 c  cn
i 1
3.
n
n
n
i 1
i 1
i 1
  ai  bi    ai   bi
Property 2: Students may need to write out and compute a few sums like this to reason
through the property. The idea is that a constant is being added to itself n times. This is
equivalent to multiplication; multiplication is repeated addition.
Property 3: One way to make sense of this property is to bring the field properties,
specifically commutative and associative properties.
c. Express each series using summation notation. Then find the sum.
i. 2 + 4 + 6 + … + 24
ii. 5 + 8 + 11 + 14 + … + 41
12
12 12  1 
2
i

2
i  2
  156


2
i 1
i 1


13
13
13
13 14  
ii.  3i  2  3 i   2  3 
  2 13  299
i 1
i 1
i 1
 2 
12
i.
d. Compute each sum using the properties of sums.
20
i.
  3i  4 
i 1
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20
ii.
 4i
i 1
20
iii.
  4i 
i 1
20
i.
20
  3i  4   3 i  4  20   3  210   80  710
i 1
ii.
iii.
i 1
20
20
i 1
20
i 1
 4i  4 i  4  210   840
20
  4i   4 i  4  210   840
i 1
i 1
Comment:
If the students have not been shown how to check their answers on the calculator, this
would be a good time to show them. Students can be required to show that they know
the properties by writing out their work, but they could be allowed to use the calculator
to check their work when possible.
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Notes on Fascinating Fractals Learning Task
This task reviews students’ understanding of geometric sequences and builds on that
knowledge to examine geometric series. Students will examine a variety of geometric
sequences that arise in fractals. They will investigate the number of triangles in different
iterations of fractals, lengths of segments, perimeters, and areas. Divergence and converge will
be discussed, along with when series diverge or converge. Students will also practice
computing geometric sums. Links are also made to summing series and horizontal asymptotes.
Supplies Needed:
 Scissors
 Cardboard or large paper
 Rulers
 Pencils and erasers
 Calculators
FASCINATING FRACTALS
Sequences and series arise in many classical mathematics problems as well as in more recently
investigated mathematics, such as fractals. The task below investigates some of the interesting
patterns that arise when investigating manipulating different figures.
Part One: Koch Snowflake4
4
Often the first picture is called stage 0. For this problem, it is called stage 1. The Sierpinski
Triangle, the next problem, presents the initial picture as Stage 0.
A number of excellent applets are available on the web for viewing iterations of fractals.
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(Images obtained from Wikimedia Commons at
http://commons.wikimedia.org/wiki/Koch_snowflake)
This shape is called a fractal. Fractals are geometric patterns that are repeated at ever smaller
increments. The fractal in this problem is called the Koch snowflake. At each stage, the middle
third of each side is replaced with an equilateral triangle. (See the diagram.)
To better understand how this fractal is formed, let’s create one!
On a large piece of paper, construct an equilateral triangle with side lengths of 9 inches.
Now, on each side, locate the middle third. (How many inches will this be?) Construct a new
equilateral triangle in that spot and erase the original part of the triangle that now forms the
base of the new, smaller equilateral triangle.
How many sides are there to the snowflake at this point? (Double-check with a partner before
continuing.)
Now consider each of the sides of the snowflake. How long is each side? Locate the middle
third of each of these sides. How long would one-third of the side be? Construct new
equilateral triangles at the middle of each of the sides.
How many sides are there to the snowflake now? Note that every side should be the same
length.
Continue the process a few more times, if time permits.
1. Now complete the first three columns of the following chart.
Stage 1
Stage 2
Stage 3
Number of
Segments
3
12
48
Length of each
Segment (in)
9
3
1
Perimeter (in)
27
36
48
2. Consider the number of segments in the successive stages.
a. Does the sequence of number of segments in each successive stage represent an arithmetic
or a geometric sequence (or neither)? Explain.
The number of segments is 4 times the number in the previous stage, so this is a
geometric sequence. In a geometric sequence, consecutive terms differ by a common
ratio. In this case, that ratio is 4.
b. What type of graph does this sequence produce? Make a plot of the stage number and
number of segments in the figure to help you determine what type of function you will use
to model this situation.
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The plot will produce an exponential graph. So the explicit formula will be the equation
of an exponential function.
c. Write a recursive and explicit formula for the number of segments at each stage.
Comment:
Students may need some guidance and a direct refresher about exponential functions to
come up with the explicit formula.
Solution:
a1  3
and an = 3(4n-1)

an  4an 1
d. Find the 7th term of the sequence. Find 12th term of the sequence. Now find the 16th. Do
the numbers surprise you? Why or why not?
a7 = 3(46) = 12,288; a12 = 3(411) = 12,582,912; a16 = 3,221,225,472. Some students may
claim that they expect the numbers to go up so fast because it’s exponential. Others may
be surprised.
3. Consider the length of each segment in the successive stages.
a. Does this sequence of lengths represent an arithmetic or a geometric sequence (or
neither)? Explain.
The number of segments is 1/3 the number in the previous stage, so this is a geometric
sequence. In a geometric sequence, consecutive terms differ by a common ratio. In this
case, that ratio is 1/3.
b. Write a recursive and explicit formula for the length of each segment at each stage.
 a1  9
n 1

1
and an  9  

1
3
 an  3 an 1
c. Find the 7th term of the sequence. Find the 12th term of the sequence. Now find the 16th.
How is what is happening to these numbers similar or different to what happened to the
sequence of the number of segments at each stage? Why are these similarities or
differences occurring?
a7 = 9(1/3)6 = 1/81  .012345679; a12 = 9(1/3)11 = 1/(39) = 1/19683 5.08 x 10-5; and a16
= 9(1/3)15 = 1/(313)  6.27 x 10-7. In the other problem, the numbers were getting bigger.
In this problem, they are getting smaller. This is because the common ratio was bigger
than one in the first problem and less than one in this problem.
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4. Consider the perimeter of the Koch snowflake.
a. How did you determine the perimeter for each of the stages in the table?
To find the perimeter, multiply the number of segments by the length of each segment.
b. Using this idea and your answers in the last two problems, find the approximate perimeters
for the Koch snowflake at the 7th, 12th, and 16th stages.
Perimeter of the 7th stage: 151.70 inches. Perimeter of the 12th stage: 639.28 inches.
Perimeter of the 16th stage: 2020.43 inches.
c. What do you notice about how the perimeter changes as the stage increases?
As we increase the stage, the perimeter gets much larger.
d. Extension: B. B. Mandelbrot used the ideas above, i.e. the length of segments and the
associated perimeters, in his discussion of fractal dimension and to answer the question,
“How long is the coast of Britain?” Research Mandelbrot’s argument and explain why
some might argue that the coast of Britain is infinitely long.
Solutions will vary depending on how this is used in the classroom.
5. Up to this point, we have not considered the area of the Koch snowflake.
a. Using whatever method you know, determine the exact area of the original triangle.
Comment:
It will be helpful for the students to recall or derive the formula for the area of an
3
equilateral triangle: A  s 2
. This would be a good place to review the derivation
4
of the formula to help students remember their right triangle trig in preparation for the
upcoming trig units.
Solution:
The area of the original triangle is
81 3
square inches.
4
b. How do you think we might find the area of the second stage of the snowflake? What
about the third stage? The 7th stage? Are we adding area or subtracting area?
Students can discuss conjectures here. Ideally, someone will suggest that we find the
area of the “new” triangles that are added at each stage, multiply by the number of
new triangles, and then add to the previous area. Students may also need to be guided
to this idea.
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c. To help us determine the area of the snowflake, complete the first two columns of the
following chart. Note: The sequence of the number of “new” triangles is represented by a
geometric sequence. Consider how the number of segments might help you determine
how many new triangles are created at each stage.
Stage
Number of
Segments
1
2
3
4
5
…
n
3
12
48
192
768
…
3(4n-1)
“New”
triangles
created
-3
12
48
192
…
3(4n-2), n > 2
Area of each of
the “new”
triangles
--
Total Area of the
New Triangles
--
d. Determine the exact areas of the “new” triangles and the total area added by their creation
for Stages 1 – 4. Fill in the chart above. (You may need to refer back to problem 1 for the
segment lengths.)
Again, using the formula for the equilateral triangle, we get the following:
Stage
Number of
“New”
Area of each of
Total Area of the
Segments
triangles
the “new”
New Triangles
created
triangles
3
---1
12
3
2
9 3
27 3
4
4
48
12
3
3
12 3
3 3
4
4
2
192
48
4
48 3 4 3
3
1 3


 
36
3
36
3 4
2
768
192
5
192 3 16 3
3
1 3


 
324
27
 9  4 324
…
…
…
3(4n-1)
3(4n-2), n > 2
n
e. Because we are primarily interested in the total area of the snowflake, let’s look at the last
column of the table. The values form a sequence. Determine if it is arithmetic or
geometric. Then write the recursive and explicit formulas for the total area added by the
new triangles at the nth stage.
This is a geometric sequence with a common ratio of 4/9.
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
81 3
27 3
n4
, a2 
a1 
4
4 and a   4 3   4 

n
 3   9 


a  4 a , n  2
 n 9 n 1
f. Determine how much area would be added at the 10th stage.
 4 3   4 6 47
16384 3
a10  
    13 3 
1594323
 3  9  3
6. Rather than looking at the area at a specific stage, we are more interested in the TOTAL
area of the snowflake. So we need to sum the areas. However, these are not necessarily
numbers that we want to try to add up. Instead, we can use our rules of exponents and
properties of summations to help us find the sum.
a. Write an expression using summation notation for the sum of the areas in the snowflake.
There may be a variety of different summations given. Based on the answer above, one
i 4
81 3 n  4 3   4 
  
possibility is
   . The important things to remember are that the
4
i 2  3   9 
first stage is not included in the summation but needs to be included in the overall
expression. Because the first stage is not included, the index of summation must start
at 2.
b. Explain how the expression you wrote in part a is equivalent to
4
i
81 3  9   4 3  n  4 
   
  .
4
 4   3  i 2  9 
Answers should address the rules of exponents and the summation properties
previously learned.
Now, the only part left to determine is how to find the sum of a finite geometric series. Let’s
take a step back and think about how we form a finite geometric series:
Sn = a1 + a1r + a1r2 + a1r3 + … + a1rn-1
Multiplying both sides by r, we get rSn = a1r + a1r2 + a1r3 ++ a1r4 … + a1rn
Subtracting these two equations:
Sn - rSn = a1 - a1rn
Factoring:
Sn(1 – r) = a1(1 – rn)
a1 1  r n
Sn 
And finally,
1 r
c. Let’s use this formula to find the total area of only the new additions through the 5th
stage. (What is a1 in this case? What is n?) Check your answer by summing the values in
your table.


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Because we are only summing 4 numbers, n = 4. Our a1 is the area added in the second
4
27 3   4  
1    
4   9   1261 3

stage. S 4 
inches2.
4
108
1
9
d. Now, add in the area of the original triangle. What is the total area of the Koch snowflake
at the fifth stage?
81 3 1261 3 862 3


inches2.
4
108
27
Do you think it’s possible to find the area of the snowflake for a value of n equal to infinity? This
is equivalent to finding the sum of an infinite geometric series. You’ve already learned that we
cannot find the sum of an infinite arithmetic series, but what about a geometric one?
e. Let’s look at an easier series: 1 + ½ + ¼ + … Make a table of the first 10 sums of this
series. What do you notice?
1
2
3
4
5
6
7
8
9
10
Terms
1
3/2
7/4
Sum
15/8 31/16 63/32 127/64 255/128 511/256 1023/512
Students should notice that the sum is approaching the value 2.
f. Now, let’s look at a similar series: 1 + 2 + 4 + …. Again, make a table. How is this table
similar or different from the one above? Why do think this is so?
1
2
3
4
5
6
7
8
9
10
Terms
1
3
7
Sum
15
31
63
127
255
511
1023
Students should notice that the sum keeps getting larger. Because the ratio is “small” in the
first problem, the sum approaches a specific number.
Recall that any real number -1 < r < 1 gets smaller when it is raised to a positive power; whereas
numbers less than -1 and greater than 1, i.e. r  1 , get larger when they are raised to a positive
power.
a1 1  r n
Thinking back to our sum formula, S n 
, this means that if r  1 , as n gets larger, rn
1 r
approaches 0. If we want the sum of an infinite geometric series, we now have
a 1  0 
a
S  1
 1 . We say that if sum of an infinite series exists—in this case, the sum of an
1 r
1 r
infinite geometric series only exists if r  1 --then the series converges to a sum. If an infinite
series does not have a sum, we say that it diverges. All arithmetic series diverge.


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g. Of the series in parts e and f, which would have an infinite sum? Explain. Find, using the
formula above, the sum of the infinite geometric series.
Part e would have a sum because the ratio is ½, which is less than 1.
S = 1/(1-1/2) = 2.
h. Write out the formula for the sum of the first n terms of the sequence you summed in part
g. Graph the corresponding function. What do you notice about the graph and the sum
you found?
Sn 

  2 1  .5 . The students


1  .5
1 1  .5n
n
should notice that the sum is the same as the
asymptote. This should make sense because
an asymptote describes the end behavior of
graphs and the infinite sum tells the sum as
the number of terms approaches infinity.
i.
Graphs and infinite series.
Write each of the following series using sigma notation. Then find the sum of the first 20
terms of the series; write out the formula. Finally, graph the function corresponding to the
sum formula for the first nth terms. What do you notice about the numbers in the series,
the function, the sum, and the graph?
2
3
1 1
1
1. 2  2    2    2    ...
3 3
 3
2. 4  4  0.6   4  0.6   4  0.6   ...
2
3
  1  20 
  1 x 
2
1

2
   
1    
i 1
n
3


1


 
  3 ; The graph of the function f x    3  
1.  2   ; S 20  
 
1
1
i 1  3 
1
1
3
3
has an asymptote at y = 3. The ratio, r = 1/3 < 1, so the geometric series diverges.
20
4 1   0.6  
  10 ; The graph of the function
2.  4  0.6  ; S 20  
1

0.6
i 1
x
4 1   0.6  

 has an asymptote at y = 10. The ratio, r = 0.6 < 1, so the
f  x 
1  0.6
geometric series diverges.
n
i 1
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j. Let’s return to the area of the Koch Snowflake. If we continued the process of creating
new triangles infinitely, could we find the area of the entire snowflake? Explain.
Yes! The ratio of the areas is less than one so the series converges. We would find the
area using the infinite sum formula and then add on the original triangle.
k. If it is possible, find the total area of the snowflake if the iterations were carried out an
infinite number of times.
27 3
27 9
35 3
then S + original =
S 4 
3
4
4
5
20
1
9
81 3 35 3  405  243 3 162 3
inches2
Total 



4
20
20
5
This problem is quite interesting: We have a finite area but an infinite perimeter!
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Part Two: The Sierpinski Triangle
(Images taken from Wikimedia Commons at
http://en.wikipedia.org/wiki/File:Sierpinski_triangle_evolution.svg.)
Another example of a fractal is the Sierpinski triangle. Start with a triangle of side length 1. This
time, we will consider the original picture as Stage 0. In Stage 1, divide the triangle into 4
congruent triangles by connecting the midpoints of the sides, and remove the center triangle. In
Stage 2, repeat Stage 1 with the three remaining triangles, removing the centers in each case.
This process repeats at each stage.
1. Mathematical Questions: Make a list of questions you have about this fractal, the Sierpinski
triangle. What types of things might you want to investigate?
Solutions will vary. The point of this question is to invite students to ask mathematical
questions. Hopefully, they will investigate some of their questions in the task.
2. Number of Triangles in the Evolution of the Sierpinski Fractal
a. How many shaded triangles are there at each stage of the evolution? How many removed
triangles are there? Use the table to help organize your answers.
Stage
Number of Shaded
Number of Newly
Triangles
Removed Triangles
1
0
0
1
3
1
2
9
3
3
27
9
4
81
27
5
243
81
b. Are the sequences above—the number of shaded triangles and the number of newly
removed triangles—arithmetic or geometric sequences? How do you know?
The two sequences are essentially the same. They are both geometric. The common
ratio is 3.
c. How many shaded triangles would there be in the nth stage? Write both the recursive and
explicit formulas for the number of shaded triangles at the nth stage.
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a0  1
and an = 3n

a

3
a
n 1
 n
d. How many newly removed triangles would there be in the nth stage? Write both the
recursive and explicit formulas for the number of newly removed triangles at the nth
stage.
a0  0, a1  3
and an = 3n-1, n>0

an  3an 1 , n  2
e. We can also find out how many removed triangles are in each evolution of the fractal.
Write an expression for the total number of removed triangles at the nth stage. Try a few
examples to make sure that your expression is correct.
Comment:
Students may present a number of different correct and incorrect conjectures. Class
discussion can help reveal ideas and misconceptions as well as different ways of
thinking. If it is not brought up by the students, this would be a good place for the
teacher to discuss how using different indices can change the expression being summed
and vice versa.
Solution:
n
1 n i
i 1
3


 3 or
3 i 1
i 1
n 1
3
i
i 0
f. Find the total number of removed triangles at the 10th stage.
S10 = 1(1-310)/(1-3) = 29,524
g. If we were to continue iterating the Sierpinski triangle infinitely, could we find the total
number of removed triangles? Why or why not. If it is possible, find the sum.
No, we cannot find the sum. This is a divergent geometric series with r > 1.
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3. Perimeters of the Triangles in the Sierpinski Fractal
a. Assume that the sides in the original triangle are one unit long. Find the perimeters of the
shaded triangles. Complete the table below.
Stage Length of a Side of Perimeter of each Number of Shaded Total Perimeter of the
a Shaded Triangle Shaded Triangle
Triangles
Shaded Triangles
1
3
1
3
0
3
1
½
3/2
9/2
2
¼
¾
9
27/4
3
1/8
3/8
27
81/8
4
1/16
3/16
81
243/16
5
1/32
3/32
243
729/32
n
n
n
n
1/(2 )
3/(2 )
3
(3n+1)/(2n)
b. Find the perimeter of the shaded triangles in the 10th stage.
Using the expression above, (311)/(210) = 177147/1024.
c. Is the sequence of values for the total perimeter arithmetic, geometric, or neither? Explain
how you know.
The sequence is geometric. The common ratio is 3/2.
d. Write recursive and explicit formulas for this sequence. In both forms, the common ratio
should be clear.
a0  3
n
3

and an  3  

3
2
an   2  an 1
 

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4. Areas in the Sierpinski Fractal
a. Assume that the length of the side of the original triangle is 1. Determine the exact area of
each shaded triangle at each stage. Use this to determine the total area of the shaded
triangles at each stage. (Hint: How are the shaded triangles at stage alike or different?)
Comments:
If the numbers seem too cumbersome, an alternate approach is to assume that the
original area is 1 in2.
Students may identify a number of different expressions that can be used in the last row
of this column. These should be explored and discussed. This exercise could serve to
reinforce students’ skill with the properties of exponents and arithmetic.
Stage
0
Length of a Side of
a Shaded Triangle
(in)
1
Area of each
Shaded Triangle
(in2)
3
4
1
½
3
1 3

 
16
2 4
2
¼
3
1 3

 
64
4 4
3
1/8
3
1 3

 
256
8 4
4
1/16
3
1 3

 
 16  4 1024
5
1/32
3
 1  3

 
4096
 32  4
n
1/(2n)
Number of Shaded
Triangles
1
Total Area of the
Shaded Triangles
(in2)
3
4
3 3
16
2
3
2
9
9 3
64
2
27
27 3
256
2
81
81 3
1024
2
243
243 3
4096
3n
3n 3
33

 
2 n 2
2
4 4
3
22 n 2
31

 
4 4
n
n
b. Explain why both the sequence of the area of each shaded triangle and the sequence of
the total area of the shaded triangles are geometric sequences. What is the common ratio
in each? Explain why the common ratio makes sense in each case.
For the sequence of the area of each shaded triangle, the ratio is ¼. That is, to get from
one stage to the next, we multiply the previous area by ¼. This makes sense because
each side length is ½ of the previous side length, and to find the area, we square the
side length. So at each stage, the area is (1/2)2 = ¼ of the area at the previous stage.
For the sequence of the total area of the shaded triangles, the ratio is ¾. The ¼ piece is
the same as described above. Because each stage has 3 times as many shaded triangles
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as the previous stage, we must also multiply by 3 times the total area from the previous
stage. Combing these leads to the common ratio of ¾.
c. Write the recursive and explicit formulas for the sequence of the area of each shaded
triangle. Make sure that the common ratio is clear in each form.

3
n
a0 
31
4
and an 

 
4 4
1
a  a
 n 4 n 1
d. Write the recursive and explicit formulas for the sequence of the total area of the shaded
triangles at each stage. Make sure that the common ratio is clear in each form.

3
n
a0 
33
4
and an 

 
4 4
a  3 a
 n 4 n 1
e. Propose one way to find the sum of the areas of the removed triangles using the results
above. Find the sum of the areas of the removed triangles in stage 5.
3
, and subtract off the area of the shaded triangles,
4
n
n
n
33
3
33
3  3 
an 

1     . For the 5th stage, we get the
  . So we get
  
4 4
4
4 4
4   4  
5
3 3 
3 781 781 3
sum of the areas of the removed triangles to be
.

1     
4   4   4 1024 4096
We can use the original area,
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f. Another way to find the sum of the areas of the removed triangles is to find the areas of
the newly removed triangles at each stage and sum them. Use the following table to help
you organize your work.
Stage Length of a Side of
Area of each
Number of Newly
Total Area of the
a Removed
Newly Removed
Removed
Newly Removed
2
Triangle (in)
Triangle (in )
Triangles
Triangles (in2)
0
0
0
0
0
2
1
½
1
3
3
1 3

 
16
16
2 4
2
2
¼
3
3 3
3
1 3

 
64
64
4 4
2
3
1/8
9
9 3
3
1 3

 
256
256
8 4
2
4
1/16
27
27 3
3
1 3

 
1024
 16  4 1024
2
5
1/32
81
81 3
3
 1  3

 
4096
4096
 32  4
n
n
n
1/(2n) = (1/2)n
3n-1
3
31
3n 1 3  1  3  3 

 
 
 
22 n 2
4 4
22 n  2
3 4  4
g. Write the explicit and recursive formulas for the area of the removed triangles at stage n.

3
n
a  0, a1 

 0
16 and a   1  3  3 

n
 
 
3 4  4
a  3 a
n
n 1

4

h. Write an expression using summation notation for the sum of the areas of the removed
triangles at each stage. Then use this formula to find the sum of the areas of the removed
triangles in stage 5.
 3   3 5  
 1     
i
i
i
n
3 n 3
3 5 3
3  4   4   
3  2343  781 3
1 3  3




 
  
  ;
  




3
12
1024
12 i 1  4  12 i1  4  12

 4096
i 1  3  4  4 
1


4




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i. Find the sum of the areas of the removed triangles at stage 20. What does this tell you
about the area of the shaded triangles at stage 20? (Hint: What is the area of the original
triangle?)
 3   3  20  
 1     
i
3 20  3 
3  4   4   
3
3

2.990... 
. The sum of the areas of the


  


3
12 i1  4  12
3
4
4
1


4




removed triangles at stage 20 is almost the entire area of the original triangle. So the
area of the shaded triangles will eventually be 0.
j. If we were to continue iterating the fractal, would the sum of the areas of the removed
triangles converge or diverge? How do you know? If it converges, to what value does it
converge? Explain in at least two ways.
The sum of the areas of the removed triangles would converge. For one, the area can
never be greater than the area of the original triangle. Secondly, the common ratio is ¾
and because r  1 , the geometric sequence converges. The sum converges to the
original area of the triangle. This could be seen by graphing the summation formula
and looking at the asymptote of the graph. (Other reasonable explanations are
possible.)
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Part Three: More with Geometric Sequences and Series
Up to this point, we have only investigated geometric sequences and series with a positive
common ratio. We will look at some additional sequences and series to better understand how
the common ratio impacts the terms of the sequence and the sum.
1. For each of the following sequences, determine if the sequence is arithmetic, geometric, or
neither. If arithmetic, determine the common difference d. If geometric, determine the
common ratio r. If neither, explain why not.
a. 2, 4, 6, 8, 10, …
d. 2, 4, 8, 16, 32, …
b. 2, -4, 6, -8, 10, …
e. 2, -4, 8, -16, 32, …
c. -2, -4, -6, -8, -10, …
f. -2, -4, -8, -16, -32, …
a. arithmetic; d = 2
b. neither; although the numbers themselves increase by two, the signs alternate. The
differences are -6, 10, -14, etc., and is therefore not common.
c. arithmetic; d = -2
d. geometric; r = 2
e. geometric; r = -2
f. geometric; r = 2
2. Can the signs of the terms of an arithmetic or geometric sequence alternate between positive
and negative? If not, explain. If so, explain when.
The signs of an arithmetic sequence cannot alternate. If the signs alternate, the difference
could not be common. The signs of a geometric sequence can alternate if the common ratio
is negative.
n
3. Write out the first 6 terms of the series
n
  2  ,   2 
i
i1
i 0
i
  1  2  . What do you
n
, and
i 1
i
i 1
notice?
n
n
  2  : -2 + 4 – 8 + 16 – 32 + 64 + …;
  2  : 1 – 2 + 4 – 8 + 16 – 32 + … ;
i1
i 0
i
  1  2  : 2 – 4 + 8 – 16 + 32 – 64
n
i 1
i
i
+ …; Some of the things students might notice
i 1
include that all three of these geometric series have alternating signs. Comparing the first
and third ones, the students should see that they can make the first or second term negative
if they write the explicit formula correctly.
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4. Write each series in summation notation and find sum of first 10 terms.
a. 1 – ½ + ¼ - 1/8 +…
b. 3 + ¾ + 3/16 + 3/64 + …
c. - 4 + 12 – 36 + 108 - …
10
 1
1   
i 1
i 1
n
10
341
 1
2
 1
a.     ;      

2
2
 1  512
i 1 
i 1 
1   
 2
  1 10 
3
 1    
i 1
i 1
n
10
4 
1
1
4
b.  3   ;  3    
1
i 1  4 
i 1  4 
1  
4
n
c.
 4  1  3
i
i 1
10
;
 4  1 3
i
i 1


4 1   3
10
  59048
1   3
Students may write different correct summation formulas. These could provide for
profitable class discourse.
i 1
i 1
5. Which of the series above would converge? Which would diverge? How do you know? For
the series that will converge, find the sum of the infinite series.
Solution: Parts a and b would converge. In both cases, we have geometric series with r  1 ,
so they converge. Part c is a geometric series with r  1 , so it diverges.

 1

 
2
i 1 
i 1
1
2


 1 3
1   
 2

and
1
3 

i 1  4 
i 1

3
1
1
4
4
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Notes on Diving into Diversions Learning Task
This task broadens the view of sequences and series beyond arithmetic and geometric with
four distinct parts.
 Students will investigate the Fibonacci sequence through the honeybee family tree
problem. The golden ratio will then be investigated, both by finding ratios and by
solving the associated quadratic. An extension links the sequence and the ratio
together to create a geometric sequence.
 Part two explores how geometric series can be applied to other mathematics,
particularly in using infinite geometric series to express repeating decimals as fractions
 Part three continues looking at different types of series, specifically series that involve
factorials. One problem will look at the infinite series used for estimating the natural
base e.
Supplies Needed:
 Calculator
 Spreadsheet capabilities (calculator or computer)
 Resources for finding examples of Fibonacci sequence and the golden ratio
DIVING INTO DIVERSIONS
Sequences and series often help us solve mathematical problems more efficiently than without
their help. Often, however, one must make conjectures, test out the conjectures, and then prove
the statements before the sequences and series can be generalized for specific situations. This
task explores some of the usefulness of sequences and series. So, let’s dive into some
mathematical diversions on sequences and series!
Part One: Fibonacci Sequence5
1. Honeybees and Family Trees
The honeybee is an interesting insect. Honeybees live in colonies called hives and they have
unusual family trees. Each hive has a special female called the queen. The other females are
worker bees; these females produce no eggs and therefore do not reproduce. The males are
called drones.
Here is what is really unusual about honeybees: Male honeybees hatch from unfertilized eggs
laid by the queen and therefore have a female parent but no male parent. Female honeybees
hatch from the queen’s fertilized eggs. Therefore, males have only 1 parent, a mother, whereas
females have both a mother and a father.
5
The traditional problem for examining the Fibonacci sequence is the rabbit problem. The
honeybee problem, however, is more realistic. A number of additional Fibonacci activities are
available on Dr. Knott’s webpage at
www.mcs.surrey.ac.uk/Personal/R.Knott/Fibonacci/fibpuzzles.html
Lessons on the Rabbit problem can be found on Dr. Knott’s page and on NCTM Illuminations.
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Unit 2
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Consider the family tree of a male honeybee.
(Picture obtained from http://www.chabad.org/media/images/112/TilE1126142.jpg on the
webpage www.chabad.org.)
a. In the first generation, there is 1 male honeybee. This male honeybee has only one parent,
a mother, at generation 2. The third generation consists of the male honeybee’s
grandparents, the mother and father of his mother. How many great-grandparents and
great-great grandparents does the male honeybee have? Explain why this makes sense.
The male honeybee will have 3 great-grandparents: his grandmother had both a mother
and a father and his grandfather had only a mother. The male honeybee has 5 greatgreat grandparents: each of his great grandmothers had 2 parents and his great
grandfather only had a mother.
b. How many ancestors does the male honeybee have at each previous generation? Complete
the following table. Find a way to determine the number of ancestors without drawing out
or counting all the honeybees.
Term
#
Value
1
2
3
4
5
6
7
8
9
10
1
1
2
3
5
8
13
21
34
55
Comment:
The students could also look at the number of male and female honeybees at each
generation.
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c. The sequence of the number of bees in each generation is known as the Fibonacci
sequence. Write a recursive formula for the Fibonacci sequence. (Hint: The recursion
formula involves two previous terms.)
F1 = ______ F2 = _______ Fn = ________ , n  2
Fn = Fn-2 + Fn-1, n  2
F1 = 1 F2 = 1
d. The Fibonacci sequence is found in a wide variety of objects that occur in nature: plants,
mollusk shells, etc. Find a picture of an item that exhibits the Fibonacci sequence and
explain how the sequence can be seen.
Solutions will vary.
2. Golden Ratio
The sequence of Fibonacci is not the only interesting thing to arise from examining the
Honeybee problem (or other similar phenomena). This next problem investigates an amazing fact
relating the Fibonacci sequence to the golden ratio.
a. Using a spreadsheet or the list capabilities on your graphing calculator to create a list of the
first 20 terms of the Fibonacci sequence and their ratios.
 In the first column, list the term number.
 In the second column, record the value of that term. (Spreadsheets can do this quickly.)
 In the third column, make a list, beginning in the second row, of each term and its
preceding term. For example, in the row with term 2, the calculate F2/F1. (Again, a
spreadsheet can do this quite quickly.
Record the ratios in the table below.
What do you notice about the ratios?
Term
2
3
4
5
6
7
8
9
10
11
Ratio
1
2
1.5
1.666667
1.6
1.625
1.615385
1.619048
1.617647
1.618182
Term
12
13
14
15
16
17
18
19
20
Ratio
1.617978
1.618056
1.618026
1.618037
1.618033
1.618034
1.618034
1.618034
1.618034
Students should notice that although the ratios oscillate above and below 1.618034,
eventually the ratios all appear to be the same.
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b. Create a plot of the term number and the ratio. What do you notice?
Ratio of Terms in Fib Sequence
2
1.8
1.6
1.4
1.2
1
Ratio
0
5
10
15
20
Students should notice that as the term number increases, the graph seems to become a
horizontal line at y = 1.618034…
c. When a sequence approaches a specific value in the long run (as the term number
approaches infinity), we say that the sequence has a limit. What appears to be the limit of
the sequence of ratios of the terms in the Fibonacci sequence?
The limit appears to be 1.618034.
d. It would be nice to know the exact value of this limit. Take a step back. Find the solutions
to the quadratic equation x2 – x – 1 = 0. Find the decimal approximation of the solutions.
Does anything look familiar?
Comment:
Part d provides a good opportunity to review how to solve quadratic equations.
Solution:
1 5
; x  1.618033989 and x  -0.6180339887. The positive
2
solution is the same as the limit of the sequence above.
x2 – x – 1 = 0  x 
e. This number is called the golden ratio and is often written as  (phi). Just like the
Fibonacci sequence, the golden ratio is present in many naturally occurring objects,
including the proportions of our bodies. Find a picture of an item that exhibits the golden
ratio and explain how the ratio is manifested.
Answers will vary. Of note: A book was recently published about how to use the golden
ratio to select appropriate clothing styles for one’s body. This might be of interest to
students who are not as interested in traditional math topics.
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f. Return to your spreadsheet and create a new column of ratios. This time, find the ratio of
a term and its following term, e.g. F3/F4. What appears to be the limit of this sequence of
ratios?
The limit of this sequence appears to be 0.618034.
g. How do you think the limit in part f might be related to the golden ratio? Test out your
conjectures and discuss with your classmates.
The limit is part f is (1) one less than the golden ratio, (2) the absolute value of the
negative solution to the above quadratic, and (3) equal to the reciprocal of the golden
ratio. What a really cool number!
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Part Two: Does 0.9999… = 1?
1. In previous courses, you may have learned a variety of ways to express decimals as fractions.
For terminating decimals, this was relatively simple. The repeating decimals, however, made
for a much more interesting problem. How might you change 0.44444444…. into a fraction?
Answers will vary depending on students’ background knowledge.
2. One way you may have expressed repeating decimals as fractions is through the use of
algebra.
a. For instance, let x = 0.44444…. What does 10x equal? You now have two equations.
Subtract the two equations and solve for x. Here’s your fraction!
If x = 0.4444…, then 10x = 4.4444….. So the difference is 9x = 4. x = 4/9.
b. Use the algebra method to express each of the following repeating decimals as fractions.
0.7777….
0.454545….
0.255555…
x = 0.7777…  10x = 7.7777…  9x = 7  x = 7/9
x = .454545…  100x = 45.4545…  99x = 45  x = 45/99 = 5/11
x = .255555…  100x = 25.5555…  99x = 25.3  x = 25.3/99 = 253/990 = 23/90
3. We can also use geometric series to express repeating decimals as fractions. Consider
0.444444… again.
a. How can we write this repeating decimal as an infinite series?
0.44444… = .4 + .04 + .004 + …
b. Now find the sum of this infinite series.
S
.4
.4 4
 
1  .1 .9 9
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4. Express each of the following repeating decimals as infinite geometric series. Then find the
sum of each infinite series.
a. 0.5
b. 0.47
c. 0.16
5
5
5
.5
5


 ... 

10 100 1000
1  .1 9
47
47
47
.47
47
b. 0.47 


 ... 

100 10, 000 1, 000, 000
1  .01 99
6
6
6
.06
1 6
1 1
5 1
c. 0.16  0.1 


 ...  0.1 
 
  

100 1000 10, 000
1  .1 10 90 10 15 30 6
a. 0.5 
5. Let’s return to the original question: does 0.9999… = 1? Use at least two different methods to
answer this question.
For the two methods discussed above, we have the following:
Method 1: x = 0.999…  10x = 9.999…  9x = 9  x = 1
9
9
.9
 ... 
 1.
Method 2: 0.999...  
10 100
1  .9
Other methods could include using a number line and estimating method, finding fractional
forms of 1/9 and 8/9 and summing those, etc.
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Unit 2
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Part Three: Factorial Fun: Lots of useful sequences and series involve factorials. You first
learned about factorials in Math 1 when you discussed the binomial theorem, combinations and
permutations.
1. As a brief refresher, calculate each of the following. Show how each can be calculated without
the use of a calculator.
5!
10!
a. 0!
b. 4!
c.
d.
3!
7!3!
a. 1
5! 5  4  3  2 1
c.

 20
3!
3  2 1
b. 4x3x2x1 = 24
10! 10  9  8  7! 10  3  3  2  4


 120
d.
7!3! 7!  3  2 1
3  2 1
2. Write out the first five terms of each of the following sequences.
2n
a. an 
 n  1!
8 4
a. 2, 4, 4, ,
3 3
3 2 5
b. 1, 2, , ,
2 3 24
 1
an 
 n  1!
n 1
n2
b. an 
n!
c.
c.
1 1 1
1
1
, , ,
,
2 6 24 120 720
3. Consider the sequences in number 2. Do you think each will diverge or converge? Explain.
(Hint: It may be helpful to plot the sequences in a graphing calculator using “Seq” mode.)
All three will converge to 0. Students could reason through this in a number of ways,
including using the graphs or appealing to the magnitudes of the numerator and
denominator as n gets larger. Students could be challenged to write divergent sequences
involving factorials. This question continues preparing students for their upcoming study of
rational functions.
4. Write an explicit formula, using factorials, for each of the following sequences.
1 1 1
1 1 3 1
a. 2, 4, 12, 40, 240, …
b. 1,1, , , ,...
c. , , , ,...
2 6 24
2 3 10 5
a. an = 2(n!)
b. an 
n
n!
c. an 
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n!
 n  1!
Georgia Department of Education
GPS Pre-Calculus
Unit 2
1st Teacher’s Edition
5. The Exciting Natural Base e
In GPS Algebra, we saw that we could approximate the transcendental number e by considering
the compound interest formula.
a. What is the formula for compounded interest and what value of e does your calculator
provide?
nt
 r
A  P  1   , where P = principal amount, r = rate, n = number of times compounded
 n
each year, and t = the number of years invested. e  2.718281828.
b. Find the value of a $1 investment at 100% for 1 year at different values of n.
N
1
2
4
12
52
100
1000
10,000
100,000
Value of Investment
2
2.25
2.37037037…
2.61303529…
2.69259695…
2.70481382…
2.71692393…
2.71814592…
2.71826823…
c. What investment value is being approached as n increased? So we say that as the value of
n increased towards infinity, the expression approaches _____.
e
d. We can also approximate e using a series with factorials. Write out and sum the first ten
n
1
terms of  . What do you notice?
i 1 i !
1 1 1 1 1 1 1 1 1 1
          2.718281526 .
0! 1! 2! 3! 4! 5! 6! 7! 8! 9!
This value is very close to e.
e. Which approximation, the one employing the compound interest formula or the one using
factorials, is more accurate with small values of n?
The factorial approximation is more accurate faster.
In the future, perhaps in calculus, you will continue encountering sequences and values that can
easily be expressed with factorials. Good luck!
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Notes on Interior Design Interview Culminating Task
This task requires students to use their knowledge of sequences and series in a specific
context. In the first two problems, the handshake and the salary situations, students must
justify which scenarios contain arithmetic or geometric sequences or series. They must then
complete a number of computations within the scenario. The final problem is much more
open-ended. The students must create three of their own designs for art pieces. One design
must have an arithmetic quality, one a geometric, and one neither. However, students should
be allowed to design fewer pieces if they can illustrate multiple types of sequence within the
single design.
Supplies Needed:
 Graph paper
 Calculators
INTERIOR DESIGN INTERVIEW
Upon graduation with a degree in interior design, Alexandra went in search of a job. One design
agency, Golden Ratio Designs, specializes in design aspects that often have a mathematical
quality. Therefore, one of the requirements for the job was a certain level of mathematical
proficiency. Alex submitted her resume’ and pictures of some of her work to the Golden Ratio
agency, she was called in for an interview.
Part One: The Handshake Problem
As part of the interview, Alex was to meet with a number of members of the agency as well as
with some of the clients. Because Alex possessed great mathematical curiosity, she wondered
how many handshakes would be shared if everyone how added the meeting shook hands with
everyone else. She did not know exactly how many people would be at the meeting, but she
wanted to come up with a way to quickly calculate the number of handshakes while she was at
the meeting. Alex hoped to impress the group with her calculation before her interview officially
began.
Find a formula for the number of handshakes that will be shared at a meeting of n people. Your
solution should address all of the following bullets but may be organized as you’d like.
 Use a table to organize your thoughts, e.g., the number of people, the number of “new”
handshakes each time a new person is added to the group, and the total number of
handshakes.
 Determine what type of function (linear, quadratic, exponential, etc.) would represent the
expression for the total number of handshakes. Explain.
 Determine if the sequence of the total number of handshakes is arithmetic, geometric, or
neither. Defend your answer.
 Determine if the sequence of the “new” handshakes as each new person is added is
arithmetic, geometric, or neither. Defend your answer.
 Prove that your formula for the total number of handshakes is true.
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The number of handshakes is
n  n  1
2
1st Teacher’s Edition
.
Number of People
“New” Handshakes
2
3
4
5
6
n
1
2
3
4
5
n–1
Total number of
handshakes
1
3
6
10
15
n  n  1
2
The expression for the total number of handshakes is quadratic because the second finite
differences are common, but not the first finite differences. Also, the graph of the number of
people and the total number of handshakes is not linear. Instead, it is shaped like the right
half of a parabola.
The sequence of the total number of handshakes is neither arithmetic nor geometric. The
numbers increase by the natural numbers. There is no common difference nor is there a
common ratio.
Total
Handshakes
Difference
Ratio
1
3
6
10
15
21
28
---
2
3
3
2
4
5/3
5
3/2
6
7/5
7
4/3
However, the sequence of the “new” handshakes is arithmetic with a common difference of 1.
The total number of handshakes is the sum of the sequence of the “new” handshakes.
Because the sequence for the “new” handshakes is the natural numbers, i.e. 1, 2, 3, …, we
n 1
need to find the sum of the natural numbers up to n – 1 or
sequence, we know that the sum is Sn 
n  a1  an 
 i . Because this is an arithmetic
i 1
. For only 1 person, there would be no
2
handshakes, so a1 = 0. For n people, there are n – 1 handshakes, so an = n – 1. Substituting
n  0  n  1 n  n  1

into the formula gives us Sn 
.
2
2
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Part Two: Salary Options
Alex’s interview went very well. Towards the end of the meeting, her potential employers asked
her to consider three salary options. Here are her choices:
1. $24,000 for the first year, with raise of $1600 per year afterwards.
2. $24,000 for the first year, with a 5% raise on that year’s salary each year
afterwards.
3. Earn $100 the first six months, $200 the second six months, $400 the third, $800
the fourth, etc.
For each of the three options:
a. Write out the first 5 terms of the sequence. Can you tell from these five terms which
option Alex should request? Explain.
b. Determine and justify if the sequence is arithmetic, geometric, or neither.
c. Write explicit and recursive formulas for the terms of the sequence.
d. Determine how much Alex would make in year 10. (Use the explicit formula for this
calculation.)
e. Determine Alex’s total salary for the first 5 years. (Write an appropriate expression using
summation notation and compute using properties of sums and formulas for sums of
series.)
a. You cannot tell which plan Alex should choose. The first two are in terms of years and the
third option is in terms of six months.
Option/Term #
1
2
3
4
5
1
24000
25600
27200
28800
30400
2
24000
25200
26460
27783
29172.50
3
100
200
400
800
1600
b. Option 1 is arithmetic with a common difference of 1600. Option 2 is geometric with
common ratio of 1.05. Option 3 is geometric with a common ratio of 2.
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c.
Option
24K; 1600 each year
Recursive
a1  24000

an  an1  1600
b1  24000

bn  1.05bn 1
c1  100

cn  2cn1
24K; 5% each year
100 first 6 months; doubles
each six months (n = number
of 6 months employed)
Explicit
an  24000  1600  n 1
bn  24000 1.05 

cn  100 2n 1
n 1



For the third option, if n represents the number of years, the sequence is also d n  300 4n 1 .
The following calculations can also be completed with this formula. However, students will
need to explain how they derived the equation.
a. To calculate how much Alex would make in year 10, calculate the tenth term for the
first two options. For the third option, we need to first calculate that she worked 10 x 2
= 20 six-month stints. So we must calculate the 20th term of this sequence.
a10  24000  1600 10 1  38400
b10  24000 1.05 
10 1
 37231.88
For Option 3, Alex earned two salaries in year 10: One for the first 6 months and one for
the second.)
c20  100 2201  52, 428,800 + c19  100 2191  26, 214, 400 = 78,643,200



   78, 643, 200
Or d10  300 4

9
b. To calculate Alex’s total salary for the first 5 years, we need to sum years 1-5 for
options 1 and 2. For option 3, we need to find the sum for terms 1 to10.
c.
5
5  24000  30400 
Option 1: 1600i  22400 
 136, 000
2
i 1
5
Option 2:
 24000 1.05
i 1
i 1
100  2  
10
Option 3:
i 1
n 1


24000 1  1.055

1  1.05
100 1  2
1 2
10
  132, 615.15
  102,300 or
 300  4  
5
i 1
i 1

300 1  45
1 4
  102,300
Rank the options in the order which would net Alex the most money over 5 years. Is there any
point at which your first option would not be the most profitable? Explain. Are all of these
options reasonable? Explain. Which option would you advise Alex to request?
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Looking only at the totals over the 5 year time frame, either of the first two options would be
acceptable. A ranking would be 1, 2, 3. But if Alex will be with Golden Ratio Designs even six
months longer, her long-term salary would be higher on option 3. However, after 5 years, the
salary becomes unreasonably high and it was unreasonably low at the beginning.
This can be easily seen by looking at the graphs of the 3 corresponding functions on the same
graph. The following is a graph of each year’s income. Another graph could be created
comparing the functions corresponding to the sum equations. (In the graph, Option 1 is red;
Option 2 is blue; Option 3 is green.)
I would advise Alex to request either option 1 or 2. Although she would eventually make lots
of money on option 3, it is an unreasonable salary scale.
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Part Three: Mathematical Modern Art
Golden Ration Designs specializes in mathematically oriented art work. Before offering Alex a
job, the owners asked her to design three art pieces with specific mathematical qualities.
1. One piece must illustrate an arithmetic sequence in some aspect of the design. The aspect
could be in the perimeter, area, decorations within the design, etc.
2. The second design must illustrate a geometric sequence.
3. The final design should also illustrate a definable sequence, but the sequence must not be
arithmetic or geometric.
Create an example of each type of design. Explain how each design was created. For each
design, explain how the given type of sequence is exhibited and show all calculations.
Answers will vary greatly. In addition to teacher assessment, students could critique each
others’ designs and calculations. The designs could also be used as the base for an
assessment. Students should be encouraged to be creative.
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Student Version: Interior Design Interview
Upon graduation with a degree in interior design, Alexandra went in search of a job. One design
agency, Golden Ratio Designs, specializes in design aspects that often have a mathematical
quality. Therefore, one of the requirements for the job was a certain level of mathematical
proficiency. Alex submitted her resume’ and pictures of some of her work to the Golden Ratio
agency, she was called in for an interview.
Part One: The Handshake Problem
As part of the interview, Alex was to meet with a number of members of the agency as well as
with some of the clients. Because Alex possessed great mathematical curiosity, she wondered
how many handshakes would be shared if everyone how added the meeting shook hands with
everyone else. She did not know exactly how many people would be at the meeting, but she
wanted to come up with a way to quickly calculate the number of handshakes while she was at
the meeting. Alex hoped to impress the group with her calculation before her interview officially
began.
Find a formula for the number of handshakes that will be shared at a meeting of n people. Your
solution should address all of the following bullets but may be organized as you’d like.
 Use a table to organize your thoughts, e.g., the number of people, the number of “new”
handshakes each time a new person is added to the group, and the total number of
handshakes.
 Determine what type of function (linear, quadratic, exponential, etc.) would represent the
expression for the total number of handshakes. Explain.
 Determine if the sequence of the total number of handshakes is arithmetic, geometric, or
neither. Defend your answer.
 Determine if the sequence of the “new” handshakes as each new person is added is
arithmetic, geometric, or neither. Defend your answer.
 Prove that your formula for the total number of handshakes is true.
Part Two: Salary Options
Alex’s interview went very well. Towards the end of the meeting, her potential employers asked
her to consider three salary options. Here are her choices:
1. $24,000 for the first year, with raise of $1600 per year afterwards.
2. $24,000 for the first year, with a 5% raise on that year’s salary each year
afterwards.
3. Earn $100 the first six months, $200 the second six months, $400 the third, $800
the fourth, etc.
For each of the three options:
Georgia Department of Education, State School Superintendent
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Georgia Department of Education
GPS Pre-Calculus
Unit 2
1st Teacher’s Edition
a. Write out the first 5 terms of the sequence. Can you tell from these five terms which
option Alex should request? Explain.
b. Determine and justify if the sequence is arithmetic, geometric, or neither.
c. Write explicit and recursive formulas for the terms of the sequence.
d. Determine how much Alex would make in year 10. (Use the explicit formula for this
calculation.)
e. Determine Alex’s total salary for the first 5 years. (Write an appropriate expression using
summation notation and compute using properties of sums and formulas for sums of
series.)
Rank the options in the order which would net Alex the most money over 5 years. Is there any
point at which your first option would not be the most profitable? Explain. Are all of these
options reasonable? Explain. Which option would you advise Alex to request?
Part Three: Mathematical Modern Art
Golden Ration Designs specializes in mathematically oriented art work. Before offering Alex a
job, the owners asked her to design three art pieces with specific mathematical qualities.
1. One piece must illustrate an arithmetic sequence in some aspect of the design. The aspect
could be in the perimeter, area, decorations within the design, etc.
2. The second design must illustrate a geometric sequence.
3. The final design should also illustrate a definable sequence, but the sequence must not be
arithmetic or geometric.
Create an example of each type of design. Explain how each design was created. For each
design, explain how the given type of sequence is exhibited and show all calculations.
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April, 2011  Page 55 of 56
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