UNIT 7
Sequences & Series
PATTERNS
EXAMPLE:
o
o
o
Suppose every student in class has a phone
conversation with every other member of the class.
What is the minimum number of calls required?
We can show telephone conversations by using the
following diagrams:
?
CONTINUED….
o
o
o
o
How many calls are necessary for two people to
have a conversation? 1 call
How many calls are necessary for everyone to talk
to everyone else in a group of three people? 3 calls
In a group of four people? 6 calls
Use a diagram to find the number of calls needed
for five people. 10 calls
CONTINUED…
o
Which of the following formulas can you use to find
the pattern for the telephone calls? C n  n  1
a.)
o
2n  3
b.)
n  n  1  5
c.)
2
Use the formula from the previous question to find
the number of calls needed for a group of 7
students.
n  n  1 7  7  1 7  6  42



 21
2
2
2
2
o
How many calls would be needed for this class?

Some patterns can be described with a sequence, or a
list of numbers.

Each number in a sequence is a term.
EXAMPLE:

Describe the pattern formed & find the next three
terms.
a.) 27, 34, 41, 48
Add 7; 55, 62, 69
b.) 243, 81, 27, 9
Divide by 3; 3, 1, 1/3
EXAMPLE:


Suppose you drop a handball from a height of 10 ft.
After the ball hits the floor, it rebounds to 85% of its
previous height.
a.) How high will the ball rebound after its fourth bounce?
Original height of the ball: 10 ft.
After 1st bounce: 85% of 10 = .85 10   8.5 ft.
After 2nd bounce: 85% of 8.5 = .85 8.5  7.225
After 3rd bounce: 85% of 7.225 = .85  7.225  6.141
After 4th bounce: 85% of 6.141 = .85  6.141  5.220
The ball will rebound about 5.2 feet after the 4th bounce.
CONTINUED…
b.) About how high will the ball rebound after the 7th
bounce?
After 4th bounce:  5.220
After 5th bounce: 85% of 5.220  .85  5.220  4.437
After 6th bounce: 85% of 4.437  .85  4.437   3.77145
After 7th bounce: 85% of 3.77145  .85  3.77145  3.206
The ball will rebound about 3.2 feet after the 7th bounce.
CONTINUED…
c.) After what bounce will the rebound height be less than
2 feet?
After 7th bounce:  3.206
After 8th bounce: 85% of 3.206  .85  3.206  2.7251
After 9th bounce: 85% of 2.7251  .85  2.7251  2.3163
After 10th bounce:85% of 2.3163  .85  2.3163  1.969
After the 10th bounce, the rebound height will be less than
2 feet.
HOMEWORK

Worksheet: Patterns
RECURSIVE FORMULA
TERMS OF A SEQUENCE
A variable, such as a, with positive integer
subscripts, can be used to represent the terms of a
sequence.
1st Term
a1
2nd Term
n-1 Term
nth Term
a2
an-1
an
(previous term)
n+1 Term
an+1
(next term)
RECURSIVE FORMULA
A recursive formula defines the terms in a
sequence by relating each term to the ones before
it.
The pattern in yesterday’s bouncing ball example was
recursive because the height of the ball after each
bounce was 85% of its previous height.
The recursive formula that describes the ball’s height
is an  0.85an 1 , where a1=10
(the first term in the sequence always needs
to be defined for a recursive sequence)
EXAMPLE:
a.) Describe the pattern that allows you to find the
next term in the sequence 2, 4, 6, 8, 10…. Write a
recursive formula for the sequence.
Add 2 to a term to find the next term
• A recursive formula is an  an 1  2, where a1= 2
•
b.) Find the sixth & seventh terms in the sequence.
Since a5 = 10, a6 = 10+2 = 12
• Since a6 = 12, a7 = 12+2 = 14
•
CONTINUED…
c.) Find the value of term a9 in the sequence.
Term a9 is the ninth term
• a9 = a8  2   a7  2   2  14  2   2  18
•
d.) Find terms a11 & a15
• a11 = a10  2   a9  2   2  18  2   2  22
•
 a  2  2  2
  a  2  2  2  2
    22  2   2   2   2
a15 = a14  2   a  2  2 
13
12
11
 30
EXAMPLE:
Write a recursive formula for each sequence. Then
find the next term.
1 1 1 1 1
,
,...
a.) , , ,
2 4 8 16 32
b.) 144, 36, 9, 9 ,...
4
Recursive Formula:
an 
1
1
an 1 , a1 
2
2
Recursive Formula:
an 
an 1 , a  144
1
4
Next Term:
1
64
Next Term:
9
16
HOMEWORK

Worksheet: Recursive Formulas
EXPLICIT FORMULAS
EXPLICIT FORMULA
A recursive formula allows us to find the value of a term
by using the preceding term.
However, we can sometimes find the value of a term
without knowing the preceding term by using the
number of the term to calculate its value.
An explicit formula expresses the nth term
in terms of n.
EXAMPLE:
The spreadsheet shows the
perimeters of squares with sides
from 1 to 6 units long.
 The numbers in each row form a sequence.

a.) For each sequence, find the next term (a7) & the twentyfifth term (a25).

Row 2: each term is the same as its subscript; therefore, a7 = 7
and a25 = 25

Row 3: each term is 4 times its subscript; therefore,
a7 = 4(7) = 28 & a25 = 4(25) = 100.
CONTINUED…
b.) Write an explicit formula
for each sequence.
Row 2:
 Row 3:

an  n
an  4n
c.) Write the first six terms in the sequence showing
the areas of the squares. Then find a20
Area: 1, 4, 9, 16, 25, 36; 400
d.) Write an explicit formula for the sequence from
part (c).
a  n2
n
HOMEWORK

Worksheet: Explicit Formulas
ARITHMETIC SEQUENCES
ARITHMETIC SEQUENCES
In an arithmetic sequence,
the difference between consecutive terms is constant.
This difference is called the common difference.
The common difference can be positive (the terms of
the sequence are increasing in value) or negative
(the terms of the sequence are decreasing in value).
EXAMPLE:
Is the given sequence arithmetic? If so, identify the
common difference.
a.) 2, 4, 8, 16,…
There is no common difference.
It is not an arithmetic sequence.
+2 +4 +8
b.) 6, 12, 18, 24,…
+6 +6 +6
There is a common difference, therefore
this is an arithmetic sequence.
The common difference is 6.
You can use an explicit formula to find the value
of the nth term of an arithmetic sequence when the
previous term is unknown.
ARITHMETIC SEQUENCE FORMULAS
Recursive Formula: a1  a given value, an  an1  d
Explicit Formula: an  a1   n  1 d
In these formulas, an is the nth term, a1 is the 1st term,
n is the number of the term, and d is the common difference.
EXAMPLE:
Suppose you participate in a bike-a-thon for charity.
• The charity starts with $1100 in donations.
• Each participant must raise at least $35 in pledges.
• What is the minimum amount of money raised if there are 75
participants?
Why the 76th
Find the 76th term of the sequence, 1100, 1135, 1170,…
•
Use the explicit formula
Substitute a1 = 1100, n =7 6,
& d = 35
an  a1   n  1 d
a76  1100   76 1 35
a76  1100  75  35
a76  1100  2625
a76  3725
term?
With 75 participants, the bike-a-thon will raise a minimum of $3,725
EXAMPLE:
•
Use the explicit formula to find the 25th term on the
sequence 5, 11, 17, 23, 29
Use the explicit formula
Substitute a1 = 5, n =25,
&d=6
an  a1   n  1 d
a25  5   25  1 6
a25  5  24  6
a25  5  144
a25  149
ARITHMETIC MEAN
The arithmetic mean of any two numbers is the
average of the two numbers
sum of two numbers
Arithmetic Mean =
2
For any three sequential terms in an arithmetic sequence,
the middle term is the arithmetic mean of the first and
third terms.
You can use arithmetic mean to find
a missing term of an arithmetic sequence.
EXAMPLE
Find the missing term of each arithmetic sequence.
a.) 84,
? , 110
Arithmetic Mean =
sum of two numbers
2
84  110
=
2
194
=
2
= 97
b.)
24, ? , 57
Arithmetic Mean =
sum of two numbers
2
24  57
=
2
81
=
2
= 40.5
HOMEWORK

Worksheet: Arithmetic Sequences
GEOMETRIC SEQUENCES
GEOMETRIC SEQUENCES
The sequence 1, 2, 4, 8,… is a geometric sequence.
How would you describe the sequence?
Multiply by 2
The sequence 900, 300, 100,… is also a geometric sequence.
How would you describe the sequence?
Divide by 3 or Multiply by 1/3
GEOMETRIC SEQUENCES
What is common between geometric sequences?
In geometric sequences, the ratio between
consecutive terms is constant.
This ratio is called the common ratio.
EXAMPLE:
Is the given sequence geometric? If so, identify the
common ratio.
a.) 5, 15, 45, 135,…
Each term is being multiplied by 3.
So, yes, it is a geometric sequence.
x3
x3
x3
b.) 15, 30, 45, 60,…
x 2 x 1.5 x 1.3
The common ratio is 3
No common ratio; therefore,
not a geometric sequence.
GEOMETRIC SEQUENCE FORMULAS
Recursive Formula: a1  a given value, an  an1 r
n 1
a

a
r
Explicit Formula:
n
1
In these formulas, an is the nth term, a1 is the 1st term,
n is the number of the term, and r is the common ratio.
EXAMPLE:
•
•
•
•
Suppose you want a reduced copy of a photograph.
The actual length of the photograph is 10 in.
The smallest size the copier can make is 64% of the original.
Find the length of the photograph after five reductions of 64%.
For five reductions, you need to find the 6th term of the geometric
sequence 10, 6.4…
Use the explicit formula
Substitute a1 = 10, n = 6,
& r = 0.64
an  a1 r n 1
a6  10 0.6461
a6  10 0.645
a6  10 0.1074
a6  1.07
After five reductions of 64%, the photograph is about 1 in. long
GEOMETRIC MEAN
You can find the geometric mean of any two
positive numbers by taking the positive square root
of the product of the two numbers.
Geometric Mean =
product of two #'s
You can use the geometric mean to find a missing
term of a geometric sequence.
EXAMPLE
Find the missing term of each geometric sequence.
a.) 12,
?, 3
G.M. = product of two #'s
b.)
5, ? , 2.8125
G.M. = product of two #'s
= 12 3
= 5 2.8125
= 36
= 14.0625
=6
= 3.75
HOMEWORK

Worksheet: Geometric Sequences
ARITHMETIC SERIES – DAY 1
SERIES
•
A series is the expression for the sum of the terms
of a sequence
Finite sequences and series have terms that you can
count individually from 1 to a whole number n.
• Infinite sequences and series continue without end.
•
Finite Sequence
Finite Series
6, 9, 12, 15, 18
6 + 9 + 12 + 15 + 18
Infinite Sequence
3, 7, 11, 15, …
Infinite Series
3 + 7 + 11 + 15 + …
EXAMPLE:
Write the related series for each finite sequence. Then evaluate the
series.
a.) 2, 11, 20, 29, 38, 47
Related Series: 2 + 11 + 20 + 29 + 38 + 47
To evaluate,
add the series
The sum of the terms of the sequence = 147
b.) 0.3, 0.6, 0.9, 1.2, 1.5, 1.8, 2.1, 2.4, 2.7, 3.0
Related Series: 0.3 + 0.6 + 0.9 + 1.2 + 1.5 + 1.8 + 2.1 + 2.4 + 2.7 + 3.0
The sum of the terms of the sequence = 16.5
c.) 100, 125, 150, 175, 200, 225
Related Series: 100 + 125 + 150 + 175 + 200 + 225
The sum of the terms of the sequence = 975
ARITHMETIC SERIES
•
An arithmetic series is a series of terms whose
terms form an arithmetic sequence.
•
When a sequence has many terms, or when you know
only the first & last terms of the sequence, you can use
a formula to evaluate the related series quickly.
SUM OF A FINITE ARITHMETIC SERIES
The sum Sn of a Finite Arithmetic Series a1 + a2 + a3 + … + an is
n
S n   a1  an 
2
where a1 is the first term, an is the nth term,
and n is the number of terms.
EXAMPLE:
xxxxx
xxxxxxx
• Several rows of cross-stitches make up the
xxxxxxxxx
green roof in the picture to the right.
xxxxxxxxxxx
a.) Find the total number of green cross-stitches xxxxxxxxxxxxx
in the roof.
The # of cross-stitches in each row form a series.
Row 1 = 5 stitches, Row 2 = 7 stitches, …
Sum of a finite
arithmetic series
Substitute n = 5, a1 = 5,
& an = 13
n
S n   a1  an 
2
5
5
S n   5  13  18   45
2
2
There are 45 stitches in the green roof.
HOMEWORK

Worksheet: Arithmetic Series – Day 1
ARITHMETIC SERIES – DAY 2
SUMMATION NOTATION
You can use the summation symbol ∑ to write a series.
Then you can use limits to indicate how many
terms you are adding.
Limits are the least and greatest integral values of n.
upper limit,
greatest
value of n
lower limit,
least
value of n
3
  5n  1
n 1
explicit formula
for the sequence
EXAMPLE:
Use summation notation to write each series for the specified
number of terms.
a.) 3 + 6 + 9 + … for 33 terms
Write the explicit formula
Substitute a1 = 3 & d = 3
n = the # of the term, so
leave that blank
distribute & simplify
upper limit
= 33
(greatest value of n)
lower limit
(least value of n)
=1
an  a1   n  1 d
an  3   n  1 3
an  3  3n  3
an  3n
33
 3n
n 1
CONTINUED…
b.) 1 + 2 + 3 + … ; n = 6
Write the explicit formula
an  a1   n  1 d
Substitute a1 = 1 & d = 1
an  1   n 11
n = the # of the term, so
leave that blank
distribute & simplify
upper limit
=6
(greatest value of n)
lower limit
(least value of n)
=1
an  1  n  1
an  n
6
n
n 1
CONTINUED…
c.) 3 + 8 + 13 + 18 … ; n = 9
Write the explicit formula
an  a1   n  1 d
Substitute a1 = 3 & d = 5
an  3   n  1 5
n = the # of the term, so
leave that blank
distribute & simplify
upper limit
=9
(greatest value of n)
lower limit
(least value of n)
=1
an  3  5n  5
an  5n  2
9
  5n  2 
n 1
EXAMPLE:
For each sum, find the number of terms, the first term, and the last
term. Then evaluate the series.
3
a.)
  5n  1
n 1
Step 1: Find the number of terms
Since the values of n are 1, 2, & 3, there are 3 terms in the series
Step 2: Find the 1st & last terms
5n  1  5 1  1  6
last term: 5n  1  5  3  1  16
1st term:
Step 3: Evaluate
Use formula
n
S n   a1  an 
2
n
3
3
S n   a1  an    6  16    22   33
2
2
2
CONTINUED…
10
b.)
  n  3
n 1
Step 1: Find the number of terms
Since the values of n are 1 thru 10, there are 10 terms in the series
Step 2: Find the 1st & last terms
1st term:
last term:
n  3  1  3  2
n  3  10  3  7
n
Step 3: Evaluate
Use formula S n   a1  an 
2
n
10
S n   a1  an   2  7   5  5 25
2
2
CONTINUED…
5
c.)
2
n

n2
Step 1: Find the number of terms
Since the values of n are 2 thru 5, there are 4 terms in the series
Step 2: Find the 1st & last terms
1st term:
n 2  22  4
last term: n 2
 52  25
Step 3: Evaluate
Use formula
n
S n   a1  an 
2
4
n
S n   a1  an    4  25   2  29   58
2
2
HOMEWORK

Worksheet: Arithmetic Series - Day 2
GEOMETRIC SERIES – DAY 1
Evaluating a Finite Geometric Series
GEOMETRIC SERIES
A geometric series is the expressions for the sum
of the terms of a geometric sequence.
SUM OF A FINITE GEOMETRIC SERIES
The sum Sn of a Finite Geometric Series
a1 + a2 + a3 + … + an, r ≠ 1, is
Sn 
a1 1  r
n

1 r
where a1 is the first term, r is the common ratio,
and n is the number of terms.
EXAMPLE:
Evaluate the series to the given term.
a.) 3 + 6 + 12 + 24…; S6
Use the formula
Substitute a1 = 3,
r = 2, & n = 6
Simplify
Sn 
S6 
a1 1  r n 
1 r
3 1  26 
1 2
3 1  64  3  63 189



1
1
1
 189
CONTINUED…
b.) -45 + 135 - 405 +…; S5
Use the formula
Substitute a1 = -45,
r = -3, & n = 5
Simplify
Sn 
S5 
a1 1  r n 
1 r
45 1  (3)5 
1  (3)
45 1  (243) 
45  244 


4
4
10980

 2745
4
EXAMPLE:
•
•
•
In March, the Floyd family starts saving for a vacation in
August, which they expect to cost $1375.
They start with $125 and each month they plan to deposit
20% more than the previous month.
Will they have enough money for their trip?
Use the formula
Substitute a1 = 125,
r = 1.2, & n = 6
Simplify
Sn 
a1 1  r n 
1 r

125 1  1.26 
1  1.2
125 1  2.986  125  1.986 


0.2
0.2
248.25

 $1,241.25
0.2
No, the Floyd family will not have enough money for their vacation.
HOMEWORK

Worksheet: Geometric Series - Day 1
GEOMETRIC SERIES – DAY 2
Evaluating an Infinite Geometric Series
INFINITE GEOMETRIC SERIES
In some cases, you can evaluate an
infinite geometric series.
When r  1, the series converges, or gets closer and
closer, to the sum S. It will eventually have a sum.
When r  1 , the series diverges, or approaches no limit.
EXAMPLE:
Decide whether each infinite geometric series diverges or
converges. State whether the series has a sum.
a.) 1 
1 1
  ...
3 9
a1 = 1 & a2 = -1/3
therefore, r = 
1
3
Since r  1, the series converges and the series has a sum

b.)  5  2 
n 1
n 1
a1 = 5(2)1-1 = 5(2)0 = 5(1)= 5
a2 = 5(2)2-1 = 5(2)1 = 5(2) = 10
Since a1 = 5 and a2 = 10,
r=2
Since r  1, the series diverges and the series does not have a sum
CONTINUED…
Decide whether each infinite geometric series diverges or
converges. State whether the series has a sum.
c.) 1 
1 1
1

 ... a1 = 1 & a2 = 1/5
therefore, r =
5 25
5
Since r  1, the series converges and the series has a sum
d.) 4  8  16  ....
a1 = 4 & a2 = 8
therefore, r = 2
Since r  1, the series diverges and the series does not have a sum
SUM OF AN INFINITE GEOMETRIC SERIES
An infinite geometric series with
r  1 converges to the sum
a1
S
1 r
Where a1 is the first term and r is the common ratio.
EXAMPLE:
Evaluate each infinite geometric series
1 1 1
a.) 1     ...
2 4 8
1
1
a1

 2
S
1 r 1 1 1
2 2
3 3 3
b.) 3     ...
2 4 8
a1

S
1 r
3
3

2
1
 1
1    1
2
 2
EXAMPLE:
•
•
The length of the outside shell of each closed chamber of a
chambered nautilus is 0.9 times the length of the larger
chamber next to it.
Estimate the total length of the outside shell for the enclosed
chambers.
The outside edge of the largest enclosed chamber is 27 mm long, so a1 = 27
Use the formula
Substitute
a1 = 27 & r = 0.9
Simplify
a1
S
1 r
27

1  0.9
27

0.1
 270
HOMEWORK

Worksheet: Geometric Series – Day 2