Lesson 11-1 Arithmetic Sequences
Lesson 11-2 Arithmetic Series
Lesson 11-3 Geometric Sequences
Lesson 11-4 Geometric Series
Lesson 11-5 Infinite Geometric Series
Lesson 11-6 Recursion and Special Sequences
Lesson 11-7 The Binomial Theorem
Lesson 11-8 Proof and Mathematical Induction
Five-Minute Check (over Chapter 10)
Main Ideas and Vocabulary
Targeted TEKS
Example 1: Find the Next Terms
Key Concept: nth Term of an Arithmetic Sequence
Example 2: Real-World Example: Find a Particular
Term
Example 3: Write an Equation for the nth Term
Example 4: Find Arithmetic Means
• Use arithmetic sequences.
• Find arithmetic means.
• sequence
• term
• arithmetic sequence
• common difference
• arithmetic means
Preparation for TEKS P.4 The student uses sequences
and series as well as tools and technology to represent,
analyze, and solve real-life problems. (A) Represent
patterns using arithmetic and geometric sequences and
series. (B) Use arithmetic, geometric, and other
sequences and series to solve real-life problems.
Find the Next Terms
Find the next four terms of the arithmetic sequence
–8, –6, –4, ….
Find the common difference d by subtracting 2
consecutive terms.
–6 – (–8) = 2 and –4 – (–6) = 2
So, d = 2.
Now add 2 to the third term of the sequence and then
continue adding 2 until the next four terms are found.
–4
–2
+2
0
+2
2
4
+2
+2
Answer: The next four terms of the sequence are –2, 0,
2, and 4.
Find the next four terms of the arithmetic sequence 5,
3, 1, ….
A. 1, 3, 5, and 7
B. –1, –3, –5, and –7
C. –2, –4, –6, and –8
D. 2, 4, 6, and 8
0%
0%
A
B
A. A
B. 0% B
C. C
C
D. D
0%
D
Find a Particular Term
CONSTRUCTION The table below shows typical
costs for a construction company to rent a crane for
one, two, three, or four months. Assuming that the
arithmetic sequence continues, how much would it
cost to rent the crane for 24 months?
Explore
Since the difference between any
two successive costs is $15,000,
the costs form an arithmetic
sequence with common difference
15,000.
Find a Particular Term
Plan
You can use the formula for the nth term of
an arithmetic sequence with a1 = 75,000 and
d = 15,000 to find a24, the cost for 24 months.
Solve
an = a1 + (n – 1)d
Formula for nth term
a24 = 75,000 + (24 –1)(15,000)
n = 24, a1 = 75,000,
d = 15,000
a24 = 420,000
Simplify.
Answer: It would cost $420,000 to rent the crane for 24
months.
Find a Particular Term
Check
You can find the term of the sequence by
adding 15,000. From Example 2 on page 623 of
your textbook, you know the cost to rent the
crane for 12 months is $240,000. So, a12
through a24 are 240,000, 255,000, 270,000,
285,000, 300,000, 315,000, 330,000, 345,000,
360,000, 375,000, 390,000, 405,000, and
420,000. Therefore, $420,000 is correct.
CONSTRUCTION The table at the
right shows typical costs for a
construction company to rent a
crane for one, two, three, or four
months. Assuming that the
arithmetic sequence continues,
how much would it cost to rent the
crane for 8 months?
A. $120,000
B. $135,000
C. $180,000
D. $220,000
1.
2.
3.
4.
A
B
C
D
A
0%
B
C
D
Write an Equation for the nth Term
Write an equation for the nth term of the arithmetic
sequence –8, –6, –4, ….
In this sequence, a1 = –8 and d = 2. Use the nth term
formula to write an equation.
an = a1 + (n – 1)d
Formula for nth term
an = –8 + (n –1)(2)
a1 = –8, d = 2
an = –8 + 2n – 2
Distributive Property
an = 2n – 10
Simplify.
Answer: An equation is an = 2n – 10.
Write an equation for the nth term of the arithmetic
sequence 5, 3, 1, ….
A. an = –2n + 3
0%
B. an = 2n + 7
C. an = 2n + 3
D. an = –2n + 7
1.
2.
3.
4.
A
A
B
C
D
B
C
D
Find Arithmetic Means
Find the three arithmetic means between 21 and 45.
You can use the nth term formula to find the common
difference. In the sequence 21, ___, ___, ___, 45, ...,
a1 = 21 and a5 = 45.
an = a1 + (n – 1)d
Formula for the nth
term
a5 = 21 + (5 –1)d
n = 5, a1 = 21
45 = 21 + 4d
a5 = 45
24 = 4d
Subtract 21 from each
side.
6=d
Divide each side by 4.
Find Arithmetic Means
Now use the value of d to find the three arithmetic means.
21
27
+6
33
+6
39
+6
Answer: The arithmetic means are 27, 33, and 39.
Check
39 + 6 = 45 
Find the three arithmetic means between 13 and 25.
A. 16, 19, 22
B. 17, 21, 25
C. 13, 17, 21
D. 15, 17, 19
0%
0%
A
B
A. A
B. 0% B
C. C
C
D. D
0%
D
Five-Minute Check (over Lesson 11-1)
Main Ideas and Vocabulary
Targeted TEKS
Key Concept: Sum of an Arithmetic Series
Example 1: Find the Sum of an Arithmetic Series
Example 2: Real-World Example: Find the First
Term
Example 3: Find the First Three Terms
Example 4: Evaluate a Sum in Sigma Notation
• Find sums of arithmetic series.
• Use sigma notation.
• series
• arithmetic series
• sigma notation
• index of summation
Preparation for TEKS P.4 The student uses sequences
and series as well as tools and technology to represent,
analyze, and solve real-life problems. (A) Represent
patterns using arithmetic and geometric sequences and
series. (B) Use arithmetic, geometric, and other
sequences and series to solve real-life problems.
Find the Sum of an Arithmetic Series
Find the sum of the first 20 even numbers, beginning
with 2.
The series is 2 + 4 + 6 + ... + 40. Since you can see that
a1 = 2, a20 = 40, and d = 2, you can use either sum
formula for this series.
Method 1
Sum formula
n = 20, a1 = 2, a20 = 40
S20 = 10(42)
Simplify.
S20 = 420
Multiply.
Find the Sum of an Arithmetic Series
Method 2
Sum formula
n = 20, a1 = 2, d = 2
S20 = 10(42)
Simplify.
S20 = 420
Multiply.
Answer: The sum of the first 20 even numbers is 420.
What is the sum of the first 15 numbers, beginning
with 1?
A. 102
B. 120
C. 136
D. 150
0%
0%
A
B
A. A
B. 0% B
C. C
C
D. D
0%
D
Find the First Term
RADIO A radio station is giving away money every
day in the month of September for a total of $124,000.
They plan to increase the amount of money given
away by $100 each day. How much should they give
away on the first day of September, rounded to the
nearest cent?
You know the values of n, Sn, and d. Use the sum formula
that contains d.
Sum formula
n = 30, d = 100
Find the First Term
124,000 = 15(2a1 + 2900)
S30 = 124,000
124,000 = 30a1 + 43,500
Distributive Property
80,500 = 30a1
2683.33 = a1
Subtract 43,500 from each
side.
Divide each side by 30
Answer: They should give away $2683.33 the first day.
GAMES A television game show gives contestants a
chance to win a total of $1,000,000 by answering 16
consecutive questions correctly. If the value of each
question is increased by $5,000, how much is the
first question worth?
A. $5,000
B. $15,000
C. $25,000
D. $50,000
0%
1.
2.
3.
4.
A
B
C
D
A
B
C
D
Find the First Three Terms
Find the first four terms of an arithmetic series in
which a1 = 14, an = 29, and Sn = 129.
Step 1
Since you know a1, an, and Sn, use
to find n.
Find the First Three Terms
Step 2
Find d.
an = a1 + (n – 1)d
21 = 14 + (6 – 1)d
15 = 5d
3=d
Step 3
Use d to determine a2, a3, and a4.
a2 = 14 + 3 or 17
a3 = 17 + 3 or 20
a4 = 20 + 3 or 23
Answer: The first four terms are 14, 17, 20, 23.
Find the first three terms of an arithmetic series in
which a1 = 11, an = 31, and Sn = 105.
A. 16, 21, 26
0%
B. 11, 16, 21
C. 11, 17, 23, 30
1.
2.
3.
4.
A
D. 17, 23, 30, 36
A
B
C
D
B
C
D
Evaluate a Sum in Sigma Notation
Evaluate
Method 1
.
Find the terms by replacing k with 3, 4, ...,
10. Then add.
Evaluate a Sum in Sigma Notation
Method 2
Since the sum is an arithmetic series, use
the formula
. There are 8
terms. a1 = 2(3) or 7, and a8 = 2(10) + 1
or 21.
Answer: The sum of the
series is 112.
Animation: Sigma Notation
Evaluate
.
A. 85
B. 95
C. 108
D. 133
0%
0%
A
B
A. A
B. 0% B
C. C
C
D. D
0%
D
Five-Minute Check (over Lesson 11-2)
Main Ideas and Vocabulary
Targeted TEKS
Example 1: Find the Next Term
Key Concept: nth Term of a Geometric Sequence
Example 2: Find a Term Given the First Term and the
Ratio
Example 3: Write an Equation for the nth Term
Example 4: Find a Term Given One Term and the Ratio
Example 5: Find Geometric Means
• Use geometric sequences.
• Find geometric means.
• geometric sequence
• common ratio
• geometric means
Preparation for TEKS P.4 The student uses sequences
and series as well as tools and technology to represent,
analyze, and solve real-life problems. (A) Represent
patterns using arithmetic and geometric sequences and
series. (B) Use arithmetic, geometric, and other
sequences and series to solve real-life problems.
Find the Next Term
MULTIPLE-CHOICE TEST ITEM What is the missing
term in the geometric sequence 324, 108, 36, 12, ___?
A 972
B 4
Read the Test Item
C 0
D –12
Find the Next Term
Solve the Test Item
Answer: B
MULTIPLE-CHOICE TEST ITEM Find the missing
term in the geometric sequence 100, 50, 25, ___.
A. 200
B. 0
C. 12.5
D. –12.5
0%
0%
A
B
A. A
B. 0% B
C. C
C
D. D
0%
D
Interactive Lab: Investigating Discrete
Sequences and Series
Find a Term Given the First Term
and the Ratio
Find the 6th term of a geometric sequence for
which a1 = –3 and r = –2.
an = a1 ● rn–1
Formula for the nth term
a6 = –3 ● (–2)6–1
n = 6, a1 = –3, r = –2
a6 = –3 ● (–32)
(–2)5 = –32
a6 = 96
Multiply.
Answer: The sixth term is 96.
What is the fifth term of a geometric sequence for
which a1 = 6 and r = 2?
A. 48
B. 96
C. 160
D. 384
0%
1.
2.
3.
4.
A
B
C
D
A
B
C
D
Write an Equation for the nth Term
Write an equation for the nth term of the geometric
sequence 5, 10, 20, 40, ….
an = a1 ● rn–1
Formula for the nth term
an = 5 ● 2n–1
a1 = 5, r = 2
Answer: An equation is an = 5 ● 2n–1.
What is an equation for the nth term of the geometric
sequence 2, 6, 18, 54, ...?
A. an = 2 ● 6n–1
0%
B. an = 54 ● 3n–1
C. an = 3 ● 2n–1
D. an = 2 ● 3
n–1
1.
2.
3.
4.
A
A
B
C
D
B
C
D
Find a Term Given One Term and the Ratio
Find the seventh term of a geometric sequence for
which a3 = 96 and r = 2.
Step 1
Find the value of a1.
an = a1rn–1
Formula for the nth term
a3 = a1(2)3–1
n = 3, r = 2
96 = a1(2)2
a3 = 96
24 = a1
Divide each side by 4.
Step 2
Find a7.
an = a1rn–1
Formula for the nth term
a7 = 24(2)7–1
n = 7, a1 = 24, r = 2
Find a Term Given One Term and the Ratio
a7 = 1536
Use a calculator.
Answer: The seventh term is 1536.
What is the sixth term of a geometric sequence for
which a4 = 27 and r = 3?
A. 3
B. 81
C. 243
D. 2187
0%
0%
A
B
A. A
B. 0% B
C. C
C
D. D
0%
D
Find Geometric Means
Find three geometric means between 3.12 and 49.92.
Use the nth term formula to find the value of r. In the
sequence 3.12, ___, ___, ___, 49.92, a1 is 3.12 and a5
is 49.92.
an = a1rn–1
Formula for the nth term
a5 = 3.12r5–1
n = 5, a1 = 3.12
49.92 = 3.12r4
a5 = 49.92
16 = r4
Divide each side by 3.12.
±2 = r
Take the fourth root of each
side.
Find Geometric Means
There are two possible common ratios, so there are two
possible sets of geometric means. Use each value of r
to find three geometric means.
r=2
r = –2
a2 = 3.12(2) or 6.24
a2 = 3.12(–2) or –6.24
a3 = 6.24(2) or 12.48
a3 = –6.24(–2) or 12.48
a4 = 12.48(2) or 24.96
a4 = 12.48(–2) or –24.96
Answer: The geometric means are 6.24, 12.48, and
24.96, or –6.24, 12.48, and –24.96.
Find three geometric means between 12 and 0.75.
A. 6, 3, 1.5 or –6, 3, –1.5
B. 6, 3, 1.5 or 6, –3, 1.5
C. 6, 3, 1.5
D. –6, 3, –1.5
0%
0%
A
B
A. A
B. 0% B
C. C
C
D. D
0%
D
Five-Minute Check (over Lesson 11-3)
Main Ideas and Vocabulary
Targeted TEKS
Key Concept: Sum of a Geometric Series
Example 1: Real-World Example: Find the Sum of the
First n Terms
Example 2: Evaluate a Sum Written in Sigma Notation
Example 3: Use the Alternate Formula for a Sum
Example 4: Find the First Term of a Series
• Find sums of geometric series.
• Find specific terms of geometric series.
• geometric series
Preparation for TEKS P.4 The student uses sequences
and series as well as tools and technology to represent,
analyze, and solve real-life problems. (A) Represent
patterns using arithmetic and geometric sequences and
series. (B) Use arithmetic, geometric, and other
sequences and series to solve real-life problems.
Find the Sum of the First n
Terms
HEALTH Contagious diseases can spread very
quickly. Suppose five people are ill during the first
week of an epidemic, and each person who is ill
spreads the disease to four people by the end of the
next week. How many people have been affected by
the illness by the end of the sixth week of the
epidemic?
This is a geometric series with a1 = 5, r = 4, and n = 6.
Sum formula
n = 6, a1 = 5, r = 4
Find the Sum of the First n
Terms
= 6825
Use a calculator.
Answer: After 6 weeks, 6825 people have been
affected by the illness.
HEALTH How many people have been affected by
the illness by the end of the eighth week of the
epidemic?
A. 9100
B. 85,000
C. 109,225
D. 174,760
0%
0%
A
B
A. A
B. 0% B
C. C
C
D. D
0%
D
Evaluate a Sum Written in Sigma Notation
Evaluate
.
Method 1
Since the sum is a geometric series, you can use the
formula
.
Sum formula
n = 12, a1 = 3, r = 2
212 = 4096
Evaluate a Sum Written in Sigma Notation
S12 = 12,285
Simplify.
Method 2
Find the terms by replacing n with 1, 2, 3, 4, 5, 6, 7, 8,
9, 10, 11, and 12. Then add.
= 3(21–1) + 3(22–1) + 3(23–1) + 3(24–1) + 3(25–1) + 3(26–1) +
3(27–1) + 3(28–1) + 3(29–1) + 3(210–1) + 3(211–1) + 3(212–1)
= 3(1) + 3(2) + 3(4) + 3(8) + 3(16) + 3(32) + 3(64) +
3(128) + 3(256) + 3(512) + 3(1024) + 3(2048)
= 3 + 6 + 12 + 24 + 48 + 96 + 192 + 384 + 768 + 1536 +
3072 + 6144
Evaluate a Sum Written in Sigma Notation
= 12,285
Answer: The sum of the series is 12,285.
Evaluate
.
A. 84
B. 200
0%
C. 484
D. 600
1.
2.
3.
4.
A
B
C
D
A
B
C
D
Use the Alternate Formula for a Sum
Find the sum of a geometric series for which
a1 = 7776, an = 6, and
.
Since you do not know the value of n, use the formula
.
Alternate sum formula
a1 = 7776, an = 6,
Use the Alternate Formula for a Sum
Simplify.
Answer: The sum of the series is 6666.
What is the sum of a geometric series for which
a1 = 6, an = 120, and r = –1.5?
A. 51.6
0%
B. 74.4
C. 120.0
1.
2.
3.
4.
A
D. 727.5
A
B
C
D
B
C
D
Find the First Term of a Series
Find a1 in a geometric series for which S8 = 765 and
r = 2.
Sum formula
S8 = 765, r = 2, and n = 8
765 =
Subtract.
765 = 255a1
Divide.
3 = a1
Divide each side by 255.
Answer: The first term of the series is 3.
Find a1 in a geometric series for which S6 = 364 and
r = 3.
A. –0.5
B. 1
C. 3
D. 4.5
0%
0%
A
B
A. A
B. 0% B
C. C
C
D. D
0%
D
Five-Minute Check (over Lesson 11-4)
Main Ideas and Vocabulary
Targeted TEKS
Key Concept: Sum of an Infinite Geometric Series
Example 1: Sum of an Infinite Geometric Series
Example 2: Infinite Series in Sigma Notation
Example 3: Write a Repeating Decimal as a Fraction
• Find the sum of an infinite geometric series.
• Write repeating decimals as fractions.
• infinite geometric series
• partial sum
• convergent series
Preparation for TEKS P.4 The student uses sequences
and series as well as tools and technology to represent,
analyze, and solve real-life problems. (C) Describe limits
of sequences and apply their properties to investigate
convergent and divergent series.
Sum of an Infinite Geometric Series
A. Find the sum of
exists.
Step 1
, if it
Find the value of r to determine if the sum
exists.
the sum does not exist.
Answer: The sum does not exist.
Sum of an Infinite Geometric Series
B. Find the sum of
, if it exists.
the sum exists.
Now use the formula for the sum of an infinite geometric
series.
Sum formula
Sum of an Infinite Geometric Series
a1 = 3, r =
Simplify.
Answer: The sum of the series is 2.
A. Find the sum of the infinite geometric series, if it
exists.
2 + 4 + 8 + 16 + ...
A. 4
B. 1
C. 2
D. no sum
0%
0%
A
B
A. A
B. 0% B
C. C
C
D. D
0%
D
B. Find the sum of the infinite geometric series, if it
exists.
A. 4
B. 2
C. 1
D. no sum
0%
0%
A
B
A. A
B. 0% B
C. C
C
D. D
0%
D
Infinite Series in Sigma Notation
Evaluate
.
Sum formula
a1 = 5, r =
Simplify.
Answer: Thus,
.
Evaluate
.
A. 6
B. 3
C.
D. no sum
0%
1.
2.
3.
4.
A
B
C
D
A
B
C
D
Write a Repeating Decimal as a Fraction
Write 0.25 as a fraction.
Method 1
Write the repeating decimal as a sum.
Sum formula
Write a Repeating Decimal as a Fraction
Subtract.
Simplify.
Write a Repeating Decimal as a Fraction
Method 2
Label the given decimal.
Repeating decimal
Multiply each side by 100.
Subtract the second
equation from the third.
Divide each side by 99.
Answer: Thus,
.
Write 0.37 as a fraction.
A.
0%
B.
C.
D.
1.
2.
3.
4.
A
B
C
D
A
B
C
D
Five-Minute Check (over Lesson 11-5)
Main Ideas and Vocabulary
Targeted TEKS
Example 1: Use a Recursive Formula
Example 2: Real-World Example: Find and Use a
Recursive Formula
Example 3: Iterate a Function
• Recognize and use special sequences.
• Iterate functions.
• Fibonacci sequence
• recursive formula
• iteration
Preparation for TEKS P.4 The student uses
sequences and series as well as tools and technology
to represent, analyze, and solve real-life problems.
(B) Use arithmetic, geometric, and other sequences
and series to solve real-life problems.
Use a Recursive Formula
Find the first five terms of the sequence in which
a1 = 5 and an + 1 = 2an + 7, n ≥ 1.
an + 1 = 2an + 7
Recursive formula
a1 + 1 = 2a1 + 7
n=1
a2 = 2(5) + 7 or 17
a2 + 1 = 2a2 + 7
a3 = 2(17) + 7 or 41
a3 + 1 = 2a3 + 7
a4 = 2(41) + 7 or 89
a4 + 1 = 2a4 + 7
a1 = 5
n=2
a2 = 17
n=3
a3 = 41
n=4
Use a Recursive Formula
an + 1 = 2an + 7
a5 = 2(89) + 7 or 185
Recursive formula
a4 = 89
Answer: The first five terms of the sequence are 5, 17,
41, 89, 185.
Find the first five terms of the sequence in which
a1 = 2 and an + 1 = 3an + 2, n ≥ 1.
A. 8, 26, 80, 242, 128
B. 2, 8, 26, 80, 242
C. 2, 6, 20, 62, 188
D. 26, 80, 242, 728, 2186
0%
0%
A
B
A. A
B. 0% B
C. C
C
D. D
0%
D
Find and Use a Recursive
Formula
A. BIOLOGY Dr. Elliott is growing cells in lab
dishes. She starts with 108 cells Monday morning
and then removes 20 of these for her experiment.
By Tuesday the remaining cells have multiplied by
1.5. She again removes 20. This pattern repeats
each day in the week. Write a recursive formula for
the number of cells Dr. Elliott finds each day before
she removes any.
Let cn represent the number of cells at the beginning of
the nth day. She takes 20 out, leaving cn – 20. The
number the next day will be 1.5 times as much. So,
cn + 1 = 1.5(cn – 20).
Answer: cn + 1 = 1.5cn – 30
Find and Use a Recursive
Formula
B. How many cells will she find on Friday morning?
On the first morning, there were 108 cells, so c1 = 108.
cn + 1 = 1.5cn – 30
Recursive formula
c1 + 1 = 1.5c1 – 30
n=1
c2 = 1.5(108) – 30 or 132
c2 + 1 = 1.5c2 – 30
c3 = 1.5(132) – 30 or 168
c3 + 1 = 1.5c3 – 30
c4 = 1.5(168) – 30 or 222
c1 = 108
n=2
c2 = 132
n=3
c3 = 168
Find and Use a Recursive
Formula
cn + 1 = 1.5cn – 30
c4 + 1 = 1.5c4 – 30
c5 = 1.5(222) – 30 or 303
Recursive formula
n=4
c4 = 222
Answer: On the fifth day, there will be 303 cells.
A. BIOLOGY Dr. Scott is growing cells in lab dishes.
She starts with 100 cells Monday morning and then
removes 30 of these for her experiment. By Tuesday
the remaining cells have doubled. She again removes
30. This pattern repeats each day in the week. Write a
recursive formula for the number of cells Dr. Scott
finds each day before she removes any.
A. cn+1 = 2cn – 60
1.
2.
3.
4.
0%
B. cn+1 = 2cn – 30
C. cn+1 = 2cn + 60
D. cn+1 = 2cn + 30
A
B
C
D
A
B
C
D
B. BIOLOGY Dr. Scott is growing cells in lab dishes.
She starts with 100 cells Monday morning and then
removes 30 of these for her experiment. By Tuesday
the remaining cells have doubled. She again removes
30. This pattern repeats each day in the week. Find
the number of cells she will find on Saturday
morning.
1.
2.
3.
4.
A. 1340
0%
B. 2270
C. 5060
D. 4130
A
B
C
D
A
B
C
D
Iterate a Function
Find the first three iterates x1, x2, and x3 of the
function f(x) = 3x – 1 for an initial value of x0 = 5.
To find the first iterate x1, find the value of the function
when x0 = 5.
x1 = f(x0)
Iterate the function.
= f(5)
x0 = 5
= 3(5) – 1 or 14
Simplify.
To find the second iterate x2, substitute x1 for x.
x2 = f(x1)
Iterate the function.
= f(14)
x1 = 14
= 3(14) – 1 or 41
Simplify.
Iterate a Function
Substitute x2 for x to find the third iterate.
x3 = f(x2)
Iterate the function.
= f(41)
x2 = 41
= 3(41) – 1 or 122
Simplify.
Answer: The first three iterates are 14, 41, and 122.
Find the first three iterates x1, x2, and x3 of the
function f(x) = 2x + 1 for an initial value of x0 = 2.
A. 3, 5, 11
0%
B. 2, 5, 11
C. 5, 11, 23
1.
2.
3.
4.
A
D. 11, 23, 47
A
B
C
D
B
C
D
Five-Minute Check (over Lesson 11-6)
Main Ideas and Vocabulary
Targeted TEKS
Example 1: Use Pascal’s Triangle
Key Concept: Binomial Theorem
Example 2: Use the Binomial Theorem
Example 3: Factorials
Key Concept: Binomial Theorem, Factorial Form
Example 4: Use a Factorial Form of the Binomial Theorem
Example 5: Find a Particular Term
• Use Pascal's triangle to expand powers of binomials.
• Use the Binomial Theorem to expand powers of
binomials.
• Pascal's triangle
• Binomial Theorem
• factorial
Preparation for TEKS P.4 The student uses sequences
and series as well as tools and technology to represent,
analyze, and solve real-life problems. (D) Apply
sequences and series to solve problems including sums
and binomial expansion.
Use Pascal’s Triangle
Expand (p + q)5.
Write row 5 of Pascal’s triangle.
1
5
10
10
5
1
Use the patterns of a binomial expansion and the
coefficients to write the expansion of (p + q)5.
(p + q)5 = 1p5q0 + 5p4q1 + 10p3q2 + 10p2q3 + 5p1q4 + 1p0q5
= p5 + 5p4q + 10p3q2 + 10p2q3 + 5pq4 + pq5
Answer: (p + q)5 = p5 + 5p4q + 10p3q2 + 10p2q3 + 5pq4 +
pq5
Expand (x + y)6.
A. x6 + 21x5y1 + 35x4y2 + 21x3y3 +
7x2y4 + y6
B. 6x5y + 15x4y2 + 20x3y3 + 15x2y4
+ 6xy5
C. x6 – 6x5y + 15x4y2 – 20x3y3 +
15x2y4 – 6xy5 + y6
6
5
4 2
3 3
D. x + 6x y + 15x y + 20x y +
15x2y4 + 6xy5 + y6
0%
0%
A
B
A. A
B. 0% B
C. C
C
D. D
0%
D
Use the Binomial Theorem
Expand (t – s)8.
The expression will have nine terms. Use the sequence
to find the coefficients
for the first five terms. Use symmetry to find the
remaining coefficients.
(t – s)8
t4(–s)4 + ... + 1t0(–s)8
Use the Binomial Theorem
= t8 – 8t7s + 28t6s2 – 56t5s3 + 70t4s4 – 56t3s5 + 28t2s6
– 8ts7 + s8
Answer: (t – s)8
= t8 – 8t7s + 28t6s2 – 56t5s3 + 70t4s4 – 56t3s5 +
28t2s6 – 8ts7 + s8
Expand (x – y)4.
A. x4 + 4x3y + 6x2y2 + 4xy3 + y4
B. 6x3y + 15x2y2 + 20xy3 + 15y4 + 6
0%
C. x4 – 4x3y + 6x2y2 – 4xy3 + y4
D. 4x4 – 4x3y + 6x2y2 – 4xy3 + 4y4
1.
2.
3.
4.
A
B
C
D
A
B
C
D
Factorials
Evaluate
.
Simplify.
Answer: 15
Evaluate
.
A. 840
0%
B. 420
C. 210
D.
1.
2.
3.
4.
A
B
C
D
A
B
C
D
Use a Factorial Form of the
Binomial Theorem
Expand (3x – y)4.
(3x – y)4
Binomial Theorem,
factorial form
Let k = 0, 1, 2, 3,
and 4.
Use a Factorial Form of the
Binomial Theorem
Answer: (3x – y)4
= 81x4 – 108x3y + 54x2y2 – 12xy3 + y4
Expand (2x + y)4.
A. 16x4 + 32x3y + 24x2y2 + 8xy3 + y4
B. 32x5 + 80x4y + 80x3y2 + 40x2y3 +
10xy4 + y5
C. 8x4 + 16x3y + 12x2y + 4xy3 + y4
3
4
0%
A
D. 32x + 64x y + 48x y + 16xy + 2y
A.
0%
B.
C.
D.
A
B
C
D
0%
0%
D
2 2
C
3
B
4
Find a Particular Term
Find the fourth term in the expansion of (a + 3b)4.
First, use the Binomial Theorem to write the expression
in sigma notation.
In the fourth term, k = 3.
k=3
Find a Particular Term
= 108ab3
Answer: 108ab3
Simplify.
Find the fifth term in the expansion of (x + 2y)6.
A. 240y4
B. 240x2y4
C. 15x2y4
D. 30x2y4
0%
0%
A
B
A. A
B. 0% B
C. C
C
D. D
0%
D
Five-Minute Check (over Lesson 11-7)
Main Ideas and Vocabulary
Targeted TEKS
Key Concept: Mathematical Induction
Example 1: Summation Formula
Example 2: Divisibility
Example 3: Counterexample
• Prove statements by using mathematical induction.
• Disprove statements by finding a counterexample.
• mathematical induction
• inductive hypothesis
Reinforcement of TEKS G.3 The student applies
logical reasoning to justify and prove mathematical
statements. (D) Use inductive reasoning to formulate a
conjecture.
Summation Formula
Prove that 1 + 3 + 5 + ... + (2n – 1) = n2.
Step 1
When n = 1, the left side of the given equation
is 2(1) – 1 or 1. The right side is 12 or 1. Thus,
the equation is true for n = 1.
Step 2
Assume 1 + 3 + 5 + ... + (2k – 1) = k2 for a
positive integer k.
Step 3
Show that the given equation is true for
n = k + 1.
1 + 3 + 5 + ... + (2k – 1) + (2(k + 1) – 1)
= k2 + (2(k + 1) – 1)
Add (2(k + 1) – 1) to each
side.
Summation Formula
= k2 + 2k + 2 – 1
Add.
= k2 + 2k + 1
Simplify.
= (k + 1)2
Factor.
The last expression is the right side of the equation to
be proved, where n has been replaced by k + 1. Thus,
the equation is true for n = k + 1.
Answer: This proves that 1 + 3 + 5 + ... + (2n – 1) = n2
is true for all positive integers n.
Prove that 2 + 4 + 6 + 8 + ... + 2n = n(n + 1). The first two steps
are shown below. What is the third step?
Step 1
When n = 1, we have 2 = 1(2), which is true. Thus, the
equation is true for n = 1.
Assume 2 + 4 + 6 + 8 + ... + 2k = k(k + 1) for a positive
integer k.
B.
Show that the given equation is
true for n = k + 1.
Show that the equation is true for
n + 1.
0%
A
C.
D.
Show that the equation is true for
n = 2.
A.
B.
0%
C.
D.
A
B
0%
C
D
0%
D
Show that the given equation is
true for k = n + 1.
C
A.
B
Step 2
Divisibility
Prove that 6n – 1 is divisible by 5 for all positive
integers n.
Step 1
When n = 1, 6n – 1 = 61 – 1 or 5. Since 5 is
divisible by 5, the statement is true for n = 1.
Step 2
Assume that 6k – 1 is divisible by 5 for some
positive integer k. This means that there is a
whole number r such that 6k – 1 = 5r.
Step 3
Show that the statement is true for n = k + 1.
6k – 1 = 5r
6k = 5r + 1
6 ● 6k = 6(5r + 1)
Inductive hypothesis
Add 1 to each side.
Multiply each side by 6.
Divisibility
6k+1 = 30r + 6
Simplify.
6k+1 – 1 = 30r + 5
Subtract 1 from each side.
6k+1 – 1= 5(6r + 1)
Factor.
Since r is a whole number, 6r + 1 is a whole number.
Therefore, 6k+1 – 1 is divided by 5.
Answer: Thus, the statement is true for n = k + 1. This
proves that 6n – 1 is divisible by 5 for all
positive integers n.
Prove that 10n – 1 is divisible by 9 for all positive
integers n. What is the last line in Step 3?
Step 1
When n = 1, 10n – 1 = 101 – 1 or 9. Since 9 is divisible
by 9, the statement is true for n = 1.
Step 2
Assume that 10k – 1 is divisible by 9 for some positive
integer k.
Step 3
Show that the statement is true for n = k + 1.
10k – 1 = 9r
10k = 9r + 1
10 ● 10k = 10(9r + 1)
10k+1 = 90r + 10
10k+1 – 1 = 90r + 9
?
Prove that 10n – 1 is divisible by 9 for all positive
integers n. What is the last line in Step 3?
A. 10k+1 – 1 = 9(10r)
B. 10k ● 101 – 1 = 90r + 9
C. 10k+1 – 1 + 1 = 90r + 9 + 1
D. 10k+1 – 1 = 9(10r + 1)
0%
1.
2.
3.
4.
A
B
C
D
A
B
C
D
Counterexample
Find a counterexample for the statement that n2 + n +
5 is always a prime number for any positive integer n.
Check the first few positive integers.
Answer: The value n = 4 is a counterexample for the
formula.
What is a counterexample for the statement that
2n – 1 is always a prime number for any positive
integer n?
A. 3
0%
B. 4
C. 5
1.
2.
3.
4.
A
D. 6
A
B
C
D
B
C
D
Five-Minute Checks
Image Bank
Math Tools
Sigma Notation
Investigating Discrete Sequences
and Series
Lesson 11-1
(over Chapter 10)
Lesson 11-2
(over Lesson 11-1)
Lesson 11-3
(over Lesson 11-2)
Lesson 11-4
(over Lesson 11-3)
Lesson 11-5
(over Lesson 11-4)
Lesson 11-6
(over Lesson 11-5)
Lesson 11-7
(over Lesson 11-6)
Lesson 11-8
(over Lesson 11-7)
To use the images that are on the
following three slides in your own
presentation:
1. Exit this presentation.
2. Open a chapter presentation using a
full installation of Microsoft® PowerPoint®
in editing mode and scroll to the Image
Bank slides.
3. Select an image, copy it, and paste it
into your presentation.
(over Chapter 10)
Find the midpoint of the line segment with
endpoints at (–4, 9) and (5, –17).
A.
B.
C.
D.
A.
B.
C.
D.
A
B
C
D
0%
0%
A
B
0%
C
0%
D
(over Chapter 10)
Find the distance between the points at (–5, 2) and
(–9, 7).
A.
0%
B.
C.
D.
1.
2.
3.
4.
A
B
C
D
A
B
C
D
(over Chapter 10)
Find the exact solution of the system of equations.
2x2 + y2 = 32
x+y=4
A.
0%
B.
C.
D.
1.
2.
3.
4.
A
B
C
D
A
B
C
D
(over Chapter 10)
Find the exact solution of the system of equations.
6x2 – y2 = 18
y2 = 17 – x2
A.
B.
C.
D.
A.
B.
C.
D.
A
B
C
D
0%
0%
A
B
0%
C
0%
D
(over Chapter 10)
Which is not the equation of a hyperbola?
A.
B. y2 = 9 + 4x2
C. (x2 – 1) – (y2 – 4) = 1
D.
0%
1.
2.
3.
4.
A
B
C
D
A
B
C
D
(over Lesson 11-1)
Find the next four terms of the arithmetic sequence
18, 11, 4, … .
A. 46, 39, 32, 25
B. 11, 4, –3, –10
C. 11, 18, 25, 32
D. –3, –10, –17, –24
A.
B.
C.
D.
A
B
C
D
0%
0%
A
B
0%
C
0%
D
(over Lesson 11-1)
Find the 12th term of an arithmetic sequence for
which a1 = –1, and d = 4.
A. 43
0%
B. 45
C. 47
D. 49
1.
2.
3.
4.
A
B
C
D
A
B
C
D
(over Lesson 11-1)
Write an equation for the nth term of the arithmetic
sequence 6, 1, –4, –9.
A. an = 5n + 11
0%
B. an = –5n + 11
C. an = –5n + 6
D. an = 5n + 6
1.
2.
3.
4.
A
B
C
D
A
B
C
D
(over Lesson 11-1)
Find the three arithmetic means in the sequence
55, ___, ___, ___, 115.
A. 67, 79, 103
B. 65, 75, 105
C. 72, 87, 98
D. 70, 85, 100
A.
B.
C.
D.
A
B
C
D
0%
0%
A
B
0%
C
0%
D
(over Lesson 11-1)
The table shows shipping fees
for delivering packages. If the
cost continues to increase at the
same rate, how much will it cost
to ship a package that weighs 18
pounds? (Hint: The cost for an
18-pound package is a9.)
1.
2.
3.
4.
A. $9.00
0%
B. $9.75
C. $15.75
D. $16.50
A
B
C
D
A
B
C
D
(over Lesson 11-1)
Find the first term and the ninth term of the
arithmetic sequence. ___, 4.5, 7, 9.5, 12, …
A. 2; 14.5
0%
B. 2.5; 22
1.
2.
3.
4.
C. 2; 22
D. 2.5; 14.5
A
B
A
B
C
D
C
D
(over Lesson 11-2)
Find Sn for the arithmetic series for which a1 = 4,
an = 16, and n = 5.
A. 100
B. 50
C. 40
D. 20
0%
0%
A
B
A. A
B. 0% B
C. C
C
D. D
0%
D
(over Lesson 11-2)
Find Sn for the arithmetic series for which a1 = 4,
an = 53, and n = 20.
A. 980
B. 570
C. 490
D. 285
0%
1.
2.
3.
4.
A
B
C
D
A
B
C
D
(over Lesson 11-2)
Find the first three terms of the arithmetic series
for which a1 = 3, an = 33, and Sn = 108.
A. 3, 9, 15
0%
B. 3, 10, 17
1.
2.
3.
4.
C. 3, 21, 39
D. 3, 12, 21
A
B
A
B
C
D
C
D
(over Lesson 11-2)
Find the sum of the arithmetic series
A. 30
B. 15
C. 5
D. 2
0%
0%
A
B
A. A
B. 0% B
C. C
C
D. D
0%
D
(over Lesson 11-2)
Find the sum of the first 10 positive odd integers.
A. 20
B. 50
C. 100
D. 200
0%
1.
2.
3.
4.
A
B
C
D
A
B
C
D
(over Lesson 11-2)
Which expression defines the arithmetic series
14 + 20 + 26 + 32 + 38 + 44 + 50?
A.
0%
B.
C.
D.
1.
2.
3.
4.
A
B
C
D
A
B
C
D
(over Lesson 11-3)
Find the next two terms of the geometric sequence
3, 12, 48, … .
A. 192, 768
B. 144, 432
C. 84, 120
D. 72, 96
0%
0%
A
B
A. A
B. 0% B
C. C
C
D. D
0%
D
(over Lesson 11-3)
Find the first five terms of the geometric sequence
for which a1 = 625 and r =
A. 390,625; 78,125; 15,625; 3125, 625
B. 625, 630, 635, 640, 645
C. 625, 600, 575, 550, 525
D. 625, 125, 25, 5, 1
1.
2.
3.
4.
A
B
C
D
A
0%
B
C
D
(over Lesson 11-3)
A. 35.2
0%
1.
2.
3.
4.
B. 29.33
C. 11
D. 5.5
A
B
A
B
C
D
C
D
(over Lesson 11-3)
Write an equation for the nth term in the geometric
sequence 216, –36, 6, … .
A. an = 216(–6)n – 1
B.
C. an = 216(–6)n
D.
A.
B.
C.
D.
A
B
C
D
0%
0%
A
B
0%
C
0%
D
(over Lesson 11-3)
A certain model automobile depreciates 20% of its
value each year. If it costs $22,800 new, what is its
value at the end of 5 years?
A. $15,328.90
0%
B. $9338.88
C. $7471.10
D. $4560.00
1.
2.
3.
4.
A
B
C
D
A
B
C
D
(over Lesson 11-3)
0%
A.
B.
C.
1.
2.
3.
4.
A
B
C
D
A
D.
B
C
D
(over Lesson 11-4)
Find Sn for the geometric series 2 + 4 + 8 + … to ten
terms.
A. 2050
B. 2046
C. 1025
D. 512
0%
0%
A
B
A. A
B. 0% B
C. C
C
D. D
0%
D
(over Lesson 11-4)
Find Sn for the geometric series
A. 3060
B. 3084
C. 1020
D. 1028
0%
1.
2.
3.
4.
A
B
C
D
A
B
C
D
(over Lesson 11-4)
Find a1 in a geometric series for which S6 = 910 and
r = –3.
A. –5
0%
B. –3.75
1.
2.
3.
4.
C. 182
D. 729
A
B
A
B
C
D
C
D
(over Lesson 11-4)
Suppose 20 rabbits live in an area and the number
of rabbits triples every six months. How many
rabbits would be there in four years? (Hint: n = 9.)
A. 98,420 rabbits
B. 131,220 rabbits
C. 131,200 rabbits
D. 196,820 rabbits
0%
0%
A
B
0%
C
0%
D
A.
B.
C.
D.
A
B
C
D
(over Lesson 11-4)
Jorge has been offered different payment plans for
a job that will take 26 weeks to complete. Which
should he choose?
A. $1000 per week
B. $0.02 the first week, each
following week will be twice
the previous week's pay
C. $2500 every other week
0%
1.
2.
3.
4.
A
B
C
D
A
D. $30,000 for the entire job
B
C
D
(over Lesson 11-5)
A. 8
B. 2.67
C. –0.5
D. –1.6
0%
0%
A
B
A. A
B. 0% B
C. C
C
D. D
0%
D
(over Lesson 11-5)
A. 16
0%
B. 12
C. –3
D. –4
1.
2.
3.
4.
A
B
C
D
A
B
C
D
(over Lesson 11-5)
Find the sum of the infinite geometric series
if it exists.
0%
A. 24
B.
C. 12
D.
1.
2.
3.
4.
A
B
C
D
A
B
C
D
(over Lesson 11-5)
Write 0.575757… as a fraction.
A.
B.
C.
D.
A.
B.
C.
D.
A
B
C
D
0%
0%
A
B
0%
C
0%
D
(over Lesson 11-5)
The first swing of a pendulum is 100 millimeters.
Then it begins to swing the other way. Each swing
afterward is 60% of the previous swing. What is the
total distance the pendulum swings before it stops?
A. 590 mm
0%
B. 250 mm
1.
2.
3.
4.
C. 160 mm
D. 140 mm
A
B
C
D
A
B
C
D
(over Lesson 11-5)
A. 0.99
B.
C.
D. 11
0%
1.
2.
3.
4.
A
B
C
D
A
B
C
D
(over Lesson 11-6)
Find the first three terms of the sequence for which
a1 = 2, an + 1 = 3an – 1.
A. 2, 7, 22
B. 2, 5, 14
C. 2, 3, 6
D. 1, 2, 5
0%
0%
A
B
A. A
B. 0% B
C. C
C
D. D
0%
D
(over Lesson 11-6)
Find the first three terms of the sequence for which
a1 = –1, an + 1 = 5an + 2.
A.
0%
B. –1, –5, –15
C. –1, 7, 37
D. –1, –3, –13
1.
2.
3.
4.
A
B
C
D
A
B
C
D
(over Lesson 11-6)
Find the first three iterates of the function
f(x) = 4x + 2 for the initial value x0 = 1.
A. 12, 26, 96
0%
B. 12, 96, 392
1.
2.
3.
4.
C. 6, 26, 106
D. 1, 6, 26
A
B
A
B
C
D
C
D
(over Lesson 11-6)
Find the first three iterates of the function
f(x) = x2 + 1 for the initial value x0 = 2.
A. 5, 26, 677
B. 2, 5, 10
C. 2, 5, 26
D. 4, 16, 256
0%
0%
A
B
A. A
B. 0% B
C. C
C
D. D
0%
D
(over Lesson 11-6)
If the rate of inflation is 3%, the cost of an item in
future years can be found by iterating the function
c(x) = 1.03x. Find the cost of a $15 CD in five years.
A. $20.15
0%
B. $17.39
C. $17.25
D. $16.88
1.
2.
3.
4.
A
B
C
D
A
B
C
D
(over Lesson 11-6)
Which of the following is the fourth term of the
sequence in which a1 = 10, an + 1 = 2an – 5?
A. 5
0%
B. 15
1.
2.
3.
4.
C. 25
D. 45
A
B
A
B
C
D
C
D
(over Lesson 11-7)
Evaluate 11!.
A. 49,001,600
B. 39,916,800
C. 30,628,800
D. 29,816,700
0%
0%
A
B
A. A
B. 0% B
C. C
C
D. D
0%
D
(over Lesson 11-7)
A. 504
B. 9
C.
D. 113,603,600
0%
1.
2.
3.
4.
A
B
C
D
A
B
C
D
(over Lesson 11-7)
Expand (p + 1)5.
A. p5 + 10p4 + 5p3 + 10p2 + 5p + 1
B. p5 + 5p4 + 10p3 + 5p2 + 10p + 1
C. p5 + 5p4 + 10p3 + 10p2 + 5p + 1
0%
1.
2.
3.
4.
A
B
C
D
D. p5 + 10p4 + 5p3 + 5p2 + 10p + 1
A
B
C
D
(over Lesson 11-7)
Expand (2x + 3y)3.
A. 8x3 + 54x2y + 36y2 + 18xy + 27y3
B. 8x3 + 36x2y + 54xy2 + 18xy + 27y3
0%
D
A
0%
A
B
0%
C
D
C
D. 8x3 + 36x2y + 54xy2 + 27y3
A.
B.
0%
C.
D.
B
C. 8x3 + 54x2y + 36xy2 + 27y3
(over Lesson 11-7)
Find the fifth term in the expansion (a + b)8.
A. 70a4b4
B. 56a3b5
C. 56a5b4
D. 14a2b6
0%
1.
2.
3.
4.
A
B
C
D
A
B
C
D
(over Lesson 11-7)
Write the expanded expression
for the volume of the cube.
A. 8x3 + 12x2 – 6x – 1
B. 8x3 – 12x2 + 6x – 1
2x1.– 1
2.
3.
4.
0%
C. 8x3 + 6x2 – 12x –1
D. 8x3 – 6x2 + 12x –1
A
B
C
D
A
B
C
D
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