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sequence is called a finite series.
SOLUTION: Study Guide and Review - Chapter 10
State whether each sentence is true or false . If
false , replace the underlined term to make a
true sentence.
1. In mathematical induction, the assumption that a
conjecture works for any particular case is called the
inductive hypothesis.
SOLUTION: In mathematical induction, the assumption that a
conjecture works for any particular case is called the
inductive hypothesis.
2. A sequence that has a set number of terms is called
an infinite sequence.
SOLUTION: A sequence that has a set number of terms is called
an finite sequence.
3. A sequence a n defined as a function of n is defined
recursively.
SOLUTION: A sequence a n defined as a function of n is defined
explicitly.
4. The step in which you show that something works
for the first case is called the inductive step in
mathematical induction.
SOLUTION: The step in which you show that something works
for the first case is called the anchor step in
mathematical induction.
5. Explicitly defined sequences give one or more of the
first few terms and then define the terms that follow
using those previous terms.
SOLUTION: Recursively defined sequences give one or more of
the first few terms and then define the terms that
follow using those previous terms.
6. The sum of the first n terms of a finite or infinite
sequence is called a finite series.
SOLUTION: The sum of the first n terms of a finite or infinite
sequence is called a finite series.
7. The difference between the terms of an arithmetic
sequence
called the
common ratio.
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SOLUTION: The difference between the terms of an arithmetic
The sum of the first n terms of a finite or infinite
sequence is called a finite series.
7. The difference between the terms of an arithmetic
sequence is called the common ratio.
SOLUTION: The difference between the terms of an arithmetic
sequence is called the common difference.
8. A geometric series is the sum of the terms of a
geometric sequence.
SOLUTION: A geometric series is the sum of the terms of a
geometric sequence.
9. If a sequence does not have a limit, it is said to
converge.
SOLUTION: If a sequence does not have a limit, it is said to
diverge.
10. The Binomial Theorem is a recursive sequence that
describes many patterns found in nature.
SOLUTION: The Fibonacci sequence is a recursive sequence that
describes many patterns found in nature.
Find the next four terms of each sequence.
11. 1, 9, 17, 25, …
SOLUTION: The terms appear to increase by 8, Check. 9 – 1= 8
17 – 9 = 8
25 – 17 = 8
The next four terms are 25 + 8 = 33
33 + 8 = 41
41 + 8 = 49
49 + 8 = 57.
12. –1, 2, 7, 14, …
SOLUTION: If we subtract each term from the term that follows,
we see a pattern.
2 – (–1) = 3
7– 2=5
14 – 7 = 7
It appears that each term is generated by adding the
next successive odd number. The next four terms
Page 1
are:
14 + 9 = 23
The next four terms are 25 + 8 = 33
33 + 8 = 41
41 + 8 = 49
Study
49 +Guide
8 = 57. and Review
- Chapter 10
–8 + (–6) + (–4) +(–2) = –20
14. S 7 of a n = −n 3
12. –1, 2, 7, 14, …
SOLUTION: SOLUTION: If we subtract each term from the term that follows,
we see a pattern.
2 – (–1) = 3
7– 2=5
14 – 7 = 7
It appears that each term is generated by adding the
next successive odd number. The next four terms
are:
14 + 9 = 23
23 + 11 = 34
34 + 13 = 47
47 + 15 = 62.
Analyze the pattern.
Find the indicated sum for each sequence.
13. fourth partial sum of a n = 2n – 10
SOLUTION: Find the first four terms of the sequence.
–1 + (–8) +( –27) + (–64) + (–125) + (–216) + (–
343) = –784
Find each sum.
–8 + (–6) + (–4) +(–2) = –20
15. SOLUTION: 14. S 7 of a n = −n 3
SOLUTION: Analyze the pattern.
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Write an explicit formula and a recursive
formula for finding the nth term of each
arithmetic sequence.
17. –6, –1, 4, …
Page 2
For a recursive formula, state the first term a 1 and
SOLUTION: then indicate that the next term is the sum of the
previous term a n – 1 and d.
Study Guide and Review - Chapter 10
Write an explicit formula and a recursive
formula for finding the nth term of each
arithmetic sequence.
17. –6, –1, 4, …
a 1 = 23, a n = a n – 1 −8
Find each sum.
19. 50th partial sum of 55 + 62 + 69 + … + 398
SOLUTION: 50th partial sum of 55 + 62 + 69 + … + 398
In this sequence, a 1 = 55 and d = 7. Find the sum of
the series.
SOLUTION: First, find the common difference.
−1 − (−6) = 5 4 − (−1) = 5 For an explicit formula, substitute a 1 = −6 and d = 5
in the nth term formula.
For a recursive formula, state the first term a 1 and
20. 37th partial sum of 9 + 3 + (–3) + …
then indicate that the next term is the sum of the
previous term a n – 1 and d.
SOLUTION: S 37 of 9 + 3 + (–3) + …
a 1 = −6, a n = a n – 1 + 5
In this sequence, a 1 = 9 and d = –6. Find the 37th
partial sum of the series.
18. 23, 15, 7, …
SOLUTION: First, find the common difference.
15 − 23 = −8 7 − 15 = −8 For an explicit formula, substitute a 1 = 23 and d = −8
in the nth term formula.
Find each sum.
For a recursive formula, state the first term a 1 and
then indicate that the next term is the sum of the
previous term a n – 1 and d.
21. SOLUTION: a 1 = 23, a n = a n – 1 −8
Find each sum.
19. 50th partial sum of 55 + 62 + 69 + … + 398
SOLUTION: 50th partial sum of 55 + 62 + 69 + … + 398
In this sequence, a 1 = 55 and d = 7. Find the sum of
the series.
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The first term of this series is –83 and the last term
is –116. The number of terms is equal to the upper
bound minus the lower bound plus one, which is 12.
Therefore, a 1 = –83, a n = –116, and n = 12. Find the
sum of the series.
Page 3
Study Guide and Review - Chapter 10
Find each sum.
21. SOLUTION: Determine the common ratio and find the next
three terms of each geometric sequence.
23. 5, 7.5, 11.25, …
SOLUTION: First, find the common ratio.
The first term of this series is –83 and the last term
is –116. The number of terms is equal to the upper
bound minus the lower bound plus one, which is 12.
Therefore, a 1 = –83, a n = –116, and n = 12. Find the
sum of the series.
The common ratio is 1.5. Multiply the third term by
1.5 to find the fourth term, and so on.
Therefore, the next three terms are 16.875, 25.3125,
and 37.96875.
24. 3 + a, –12 – 4a, 48 + 16a, …
SOLUTION: First, find the common ratio.
22. SOLUTION: The first term of this series is 46 and the last term is
122. The number of terms is equal to the upper
bound minus the lower bound plus one, which is 20.
Therefore, a 1 = 46, a n = 122, and n = 20. Find the
sum of the series.
The common ratio is –4. Multiply the third term by –
4 to find the fourth term, and so on.
–4(48 + 16a) = –192–64a
–4(–192–64a) = 768 + 256a
–4(768 + 256a) = –3072 – 1024a
Therefore, the next three terms are –192 – 64a, 768
+ 256a, and –3072 – 1024a.
Determine the common ratio and find the next
three terms of each geometric sequence.
23. 5, 7.5, 11.25, …
SOLUTION: First, find the common ratio.
The common ratio is 1.5. Multiply the third term by
1.5 to find the fourth term, and so on.
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Write an explicit formula and a recursive
formula for finding the nth term of each
geometric sequence.
25. 10, –20, 40, …
SOLUTION: First, find the common ratio.
–20 ÷ 10 = –2
40 ÷ –20 = –2
Page 4
For an explicit formula, substitute a 1 = 10 and r = –2
–4(–192–64a) = 768 + 256a
–4(768 + 256a) = –3072 – 1024a
indicate that the next term is the product of the first
term a n – 1 and r.
Therefore,
nextReview
three terms
are –192 – 64a,
Study
Guidetheand
- Chapter
10 768
+ 256a, and –3072 – 1024a.
Write an explicit formula and a recursive
formula for finding the nth term of each
geometric sequence.
25. 10, –20, 40, …
a n = 162
; a 1 = 162, a n =
an – 1
Find each sum.
27. first seven terms of 80 + 32 +
+ …
SOLUTION: SOLUTION: First, find the common ratio.
First, find the common ratio.
–20 ÷ 10 = –2
40 ÷ –20 = –2
32 ÷ 80 =
÷ 32 =
For an explicit formula, substitute a 1 = 10 and r = –2
in the nth term formula.
The common ratio is
. Use Formula 1 for the sum
of a finite geometric series.
For a recursive formula, state the first term a 1. Then
indicate that the next term is the product of the first
term a n – 1 and r.
n –1
a n = 10(–2)
; a 1 = 10, a n = (–2)a n – 1
26. 162, 54, 18, …
28. a 1 = 11, a n = 360,448, r = 8
SOLUTION: First, find the common ratio.
SOLUTION: 54 ÷ 162 =
Use Formula 2 for the nth partial sum of a geometric
series.
18 ÷ 54 =
For an explicit formula, substitute a 1 = 162 and r =
in the nth term formula.
Use mathematical induction to prove that each
conjecture is true for all positive integers n.
29. 6n – 9 is divisible by 3.
SOLUTION: For a recursive formula, state the first term a 1. Then
Let Pn be the statement 6 – 9 is divisible by 3. P1 is
indicate that the next term is the product of the first
term a n – 1 and r.
the statement that 6 – 9 is divisible by 3. P1 is true
a n = 162
; a 1 = 162, a n =
an – 1
n
1
1
because 6 – 9 is –3, which is divisible by 3. Assume
Pk is true, where k is a positive integer, and show
k
that Pk + 1 must be true. That is, show that 6 – 9 =
Find each sum.
27. first seven terms of 80 + 32 +
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SOLUTION: First, find the common ratio.
k +1
+ …
3r for some integer r implies that 6
divisible by 3.
– 9 is
Page 5
n
k
implies Pk + 1, Pn is true for n = 2, n = 3, and so on.
n
Study Guide and Review - Chapter 10
Use mathematical induction to prove that each
conjecture is true for all positive integers n.
By the principle of mathematical induction, 6 – 9 is
divisible by 3 for all positive integers n.
30. 7n – 5 is divisible by 2.
SOLUTION: 29. 6n – 9 is divisible by 3.
n
Let Pn be the statement 7 – 5 is divisible by 2. P1 is
SOLUTION: 1
n
Let Pn be the statement 6 – 9 is divisible by 3. P1 is
1
the statement that 6 – 9 is divisible by 3. P1 is true
1
because 6 – 9 is –3, which is divisible by 3. Assume
Pk is true, where k is a positive integer, and show
k
that Pk + 1 must be true. That is, show that 6 – 9 =
k +1
3r for some integer r implies that 6
divisible by 3.
the statement that 7 – 5 is divisible by 2. P1 is true
1
because 7 – 5 is 2, which is divisible by 2. Assume
Pk is true, where k is a positive integer, and show
n
that Pk + 1 must be true. That is, show that 7 – 5 =
k +1
2r for some integer r implies that 7
divisible by 2.
– 5 is
– 9 is
Because r is an integer, 6r + 15 is an integer, and 3
k +1
(6r + 15) is divisible by 3. Therefore, 6
– 9 is
divisible by 3. Because Pn is true for n = 1 and Pk
implies Pk + 1, Pn is true for n = 2, n = 3, and so on.
n
By the principle of mathematical induction, 6 – 9 is
divisible by 3 for all positive integers n.
Because r is an integer, 7r + 15 is an integer, and 2
k +1
(7r + 15) is divisible by 2. Therefore, 7
– 5 is
divisible by 2. Because Pn is true for n = 1 and Pk
implies Pk + 1, Pn is true for n = 2, n = 3, and so on.
n
By the principle of mathematical induction, 7 – 5 is
divisible by 2 for all positive integers n.
31. 5n + 3 is divisible by 4.
SOLUTION: 30. 7n – 5 is divisible by 2.
n
Let Pn be the statement 5 + 3 is divisible by 4. P1 is
SOLUTION: 1
n
Let Pn be the statement 7 – 5 is divisible by 2. P1 is
1
the statement that 7 – 5 is divisible by 2. P1 is true
1
because 7 – 5 is 2, which is divisible by 2. Assume
Pk is true, where k is a positive integer, and show
n
that Pk + 1 must be true. That is, show that 7 – 5 =
k +1
2r for some integer r implies that 7
divisible by 2.
1
because 5 + 3 is 8, which is divisible by 4. Assume
Pk is true, where k is a positive integer, and show
n
that Pk + 1 must be true. That is, show that 5 + 3 =
k +1
4r for some integer r implies that 5
divisible by 4.
+ 3 is
– 5 is
Because r is an integer, 7r + 15 is an integer, and 2
k +1
(7r + 15) is divisible by 2. Therefore, 7
– 5 is
divisible by 2. Because Pn is true for n = 1 and Pk
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the statement that 5 + 3 is divisible by 4. P1 is true
Because r is an integer, 5r – 3 is an integer, and 4(5r
k +1
– 3) is divisible by 4. Therefore, 5
+ 3 is divisible
by 4. Because Pn is true for n = 1 and Pk implies Pk
+ 1, Pn
Page 6
is true for n = 2, n = 3, and so on. By the
n
principle of mathematical induction, 5 + 3 is divisible
divisible by 2. Because Pn is true for n = 1 and Pk
by 4. Because Pn is true for n = 1 and Pk implies Pk
+ 1, Pn is true for n = 2, n = 3, and so on. By the
implies Pk + 1, Pn is true for n = 2, n = 3, and so on.
n
n
By the
principle
of mathematical
induction, 710
– 5 is
Study
Guide
and
Review - Chapter
principle of mathematical induction, 5 + 3 is divisible
by 4 for all positive integers n.
divisible by 2 for all positive integers n.
31. 5n + 3 is divisible by 4.
32. 2 ⋅ 3 + 4 ⋅ 5 + 6 ⋅ 7 + … + 2n(2n + 1) =
SOLUTION: n
Let Pn be the statement 5 + 3 is divisible by 4. P1 is
SOLUTION: 1
the statement that 5 + 3 is divisible by 4. P1 is true
Let Pn be the statement 2 ⋅ 3 + 4 ⋅ 5 + 6 ⋅ 7 + … +
1
because 5 + 3 is 8, which is divisible by 4. Assume
Pk is true, where k is a positive integer, and show
2n(2n + 1) =
n
that Pk + 1 must be true. That is, show that 5 + 3 =
k +1
4r for some integer r implies that 5
divisible by 4.
is a . Because 2(1)[2(1) + 1] =
+ 3 is
true statement, Pn is true for n = 1. Assume that 2 ⋅
3 + 4 ⋅ 5 + 6 ⋅ 7 + … + 2k(2k + 1) =
is true for a positive integer k.
Show that Pk + 1 must be true.
Because r is an integer, 5r – 3 is an integer, and 4(5r
k +1
– 3) is divisible by 4. Therefore, 5
+ 3 is divisible
by 4. Because Pn is true for n = 1 and Pk implies Pk
+ 1, Pn
This final statement is exactly Pk + 1, so Pk + 1 is
true. Because Pn is true for n = 1 and Pk implies Pk
is true for n = 2, n = 3, and so on. By the
n
principle of mathematical induction, 5 + 3 is divisible
by 4 for all positive integers n.
+ 1, Pn
is true for n = 2, n = 3, and so on. That is, by
the principle of mathematical induction, 2 ⋅ 3 + 4 ⋅
32. 2 ⋅ 3 + 4 ⋅ 5 + 6 ⋅ 7 + … + 2n(2n + 1) =
5 + 6 ⋅ 7 + … + 2n(2n + 1) =
is true for all positive integers n.
SOLUTION: Let Pn be the statement 2 ⋅ 3 + 4 ⋅ 5 + 6 ⋅ 7 + … +
33. + … +
+
= 2n(2n + 1) =
. Because 2(1)[2(1) + 1] =
is a true statement, Pn is true for n = 1. Assume that 2 ⋅
3 + 4 ⋅ 5 + 6 ⋅ 7 + … + 2k(2k + 1) =
is true for a positive integer k.
Show that Pk + 1 must be true.
SOLUTION: + Let Pn be the statement
= +
is a true statement, Pn
is true for n = 1. Assume that
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This final statement is exactly Pk + 1, so Pk + 1 is
true. Because Pn is true for n = 1 and Pk implies Pk
+ …
. Because
= + … +
+ = + + is
Page 7
true for a positive integer k. Show that Pk + 1 must
be true.
+1
n
+ the principle of mathematical induction, 2 ⋅ 3 + 4 ⋅
5 + 6 ⋅ 7 + … + 2n(2n + 1) =
+ … +
is Study Guide and Review - Chapter 10
= is true for all positive integers n.
true for all positive integers n.
33. + … +
+
Prove each inequality for the indicated values
of n.
= 34. 4n
4n, for all positive integers n
SOLUTION: n
4n, for all positive
Let Pn be the statement 4
SOLUTION: 1
+ Let Pn be the statement
+ + …
integers n. P1 is true, since 4 ≥ 4(1) is a true
k
inequality. Assume Pk is true, that 4 ≥ 4k if k is a
= +
. Because
positive integer, and show that Pk + 1 must be true.
k
= is a true statement, Pn
+ is true for n = 1. Assume that
+ … +
+ = is
k +1
That is, show that 4 ≥ 4k implies 4
≥ 4(k + 1). Use both parts of the inductive hypothesis.
Note: If 12k 12 and 12 4, then by the Transitive Property of Inequality, 12k 4. k
4 ≥ 4k k
4 · 4 ≥ 4 · 4k
true for a positive integer k. Show that Pk + 1 must
be true.
k +1
4
16k
≥
k
1
12k
12
12k
4
4k + 12k 4k + 4 16k 4(k + 1)
k +1
By the Transitive Property of Inequality, if 4
≥
k +1
This final statement is exactly Pk + 1, so Pk + 1 is
true. Because Pn is true for n = 1 and Pk implies Pk
+ 1, Pn is true for n = 2, n = 3, and so on. That is, by
the principle of mathematical induction,
+ + … +
+ = 16k and 16k 4(k + 1), then 4
≥ 4(k + 1).
Since Pn is true for n = 1 and Pk implies Pk + 1 for k
1, Pn is true for n = 2, n = 3, and so on. By the
n
principle of mathematical induction, 4
positive integers n.
4n, for all
35. 5n < 6n, for all positive integers n
SOLUTION: n
is true for all positive integers n.
Let Pn be the statement 5n < 6 . P1 is true, since 5
1
(1) < 6 is a true statement. Assume Pk is true, that
k
Prove each inequality for the indicated values
of n.
34. 4n
4n, for all positive integers n
n
4n, for all positive
1
integers n. P1 is true, since 4 ≥ 4(1) is a true
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inequality.
P Cognero
is true,
k
k
that 4 ≥ 4k if k is a
positive integer, and show that Pk + 1 must be true.
k
k
1
must be true. That is, show that 5k < 6 implies 5(k
k +1
SOLUTION: Let Pn be the statement 4
5k < 6 if k is a positive integer, and show that Pk +
k +1
+ 1) < 6
. Use both parts of the inductive
hypothesis.
Note: If 5 25 and 25 25k, then by the
Transitive Property of Inequality, 5 25k.
k
5k < 6
6 ⋅ 5k < 6
k
k
Page 8
1
1 k
n
k
k +1
1, Pn is true for n = 2, n = 3, and so on. By the
n
principle of mathematical induction, 4
4n, for all
Study
Guide
andn. Review - Chapter 10
positive
integers
35. 5n < 6n, for all positive integers n
36. (4x + 6)5
SOLUTION: n
Let Pn be the statement 5n < 6 . P1 is true, since 5
5
1
(1) < 6 is a true statement. Assume Pk is true, that
k
5k < 6 if k is a positive integer, and show that Pk +
k
must be true. That is, show that 5k < 6 implies 5(k
k
4
1
3
2
2
3
1
Step 2 The numbers in the 5th row of Pascal’s
triangle are 1, 5, 10, 10, 5, and 1. Use these numbers
as the coefficients of the terms in the series. Then
simplify.
1
1 k
25 ⋅ 1
+1
0
(4x) (6) + (4x) (6) + (4x) (6) + (4x) (6) + (4x)
4
0 5
(6) +(4x) (6)
k
Step 1 Write the series for (a + b) , omitting the
coefficients and replacing a with 4x and b with 6.
The series has 5 + 1 or 6 terms. Exponents of 4x
decrease from 5 to 0. Exponents of 6 increase from
0 to 5.
5
k +1
+ 1) < 6
. Use both parts of the inductive
hypothesis.
Note: If 5 25 and 25 25k, then by the
Transitive Property of Inequality, 5 25k.
5k < 6
6 ⋅ 5k < 6
k
⋅6
k
30k < 6
n
principle of mathematical induction, 5n < 6 is true
for all positive integers n.
Use Pascal’s triangle to expand each binomial.
SOLUTION: 1
Since Pn is true for k ≥ 1 and Pk implies Pk + 1 for k
1, Pn is true for n = 2, n = 3, and so on. By the
37. (m – 5n)6
25 ⋅ k
SOLUTION: 25 25k
5 25k
5k + 5 25k + 5k 5(k + 1) 30k
6
By the Transitive Property of Inequality, if 5(k + 1)
k +1
k +1
, then 5(k + 1) < 6
.
30k and 30k < 6
Since Pn is true for k ≥ 1 and Pk implies Pk + 1 for k
1, Pn is true for n = 2, n = 3, and so on. By the
n
principle of mathematical induction, 5n < 6 is true
for all positive integers n.
Use Pascal’s triangle to expand each binomial.
Step 1 Write the series for (a + b) , omitting the
coefficients and replacing a with m and b with −5n.
The series has 6 + 1 or 7 terms. Exponents of m
decrease from 6 to 0. Exponents of −5n increase
from 0 to 6.
6
0
5
1
4
2
3
(m) (−5n) + (m) (−5n) + (m) (−5n) + (m) (−5n)
2
4
1
5
0
6
+ (m) (−5n) + (m) (−5n) + (m) (−5n)
3
Step 2 The numbers in the 6th row of Pascal’s
triangle are 1, 6, 15, 20, 15, 6, and 1. Use these
numbers as the coefficients of the terms in the
series. Then simplify.
36. (4x + 6)5
Find the coefficient of the indicated term in
each expansion.
SOLUTION: 5
Step 1 Write the series for (a + b) , omitting the
coefficients and replacing a with 4x and b with 6.
The series has 5 + 1 or 6 terms. Exponents of 4x
decrease from 5 to 0. Exponents of 6 increase from
0 to 5.
5
0
4
1
3
2
2
3
(4x) (6) + (4x) (6) + (4x) (6) + (4x) (6) + (4x)
4
0 5
(6) +(4x) (6)
1
Step 2 The numbers in the 5th row of Pascal’s
triangle are 1, 5, 10, 10, 5, and 1. Use these numbers
as the coefficients of the terms in the series. Then
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simplify.
38. (6x – 3y)10, x4y 6 term
SOLUTION: 10
n
For (6x – 3y) to have the form (a + b) , let a = 6x
and b = −3y. The coefficient of the term containing
n –r
r
n
a
b in the expansion (a + b) is given by nCr.
So, to find the binomial coefficient of the term
4 6
10
containing a b in the expansion of (a + b) ,
evaluate nCr for n = 10 and r = 6.
Page 9
Step 2 The numbers in the 6th row of Pascal’s
triangle are 1, 6, 15, 20, 15, 6, and 1. Use these
numbers as the coefficients of the terms in the
series. Then simplify.
Study Guide and Review - Chapter 10
Find the coefficient of the indicated term in
each expansion.
10
expansion of (6x – 3y)
is 198,404,640.
39. (2y + 3)13, 8th term
SOLUTION: 38. (6x – 3y)10, x4y 6 term
First find the coefficient of the 8th term in the
SOLUTION: 13
10
n
For (6x – 3y) to have the form (a + b) , let a = 6x
and b = −3y. The coefficient of the term containing
n –r
4 6
Therefore, the coefficient of the x y term in the
r
expansion of (a + b) , evaluate nCr for n = 13 and r
= 8 – 1 or 7.
n
a
b in the expansion (a + b) is given by nCr.
So, to find the binomial coefficient of the term
4 6
10
containing a b in the expansion of (a + b) ,
evaluate nCr for n = 10 and r = 6.
Thus, the binomial coefficient of the 8th term in (a +
b)
13
13
is 1716. For (2y + 3) to have the form (a + b)
13
let a = 2y and b = 3. Substitute 2y for a and 3 for
b to find the coefficient of the 8th term in the original
binomial expansion.
4 6
Thus, the binomial coefficient of the a b term in the
10
(a + b) is 210. Substitute 6x for a and −3y for b to
4 6
find the coefficient of the x y term in the original
binomial expansion.
4 6
Therefore, the coefficient of the x y term in the
10
expansion of (6x – 3y)
is 198,404,640.
Therefore, the coefficient of the 8th term in the
expansion of (2y + 3)
13
is 240,185,088.
Use the Binomial Theorem to expand each
binomial.
40. (2p 2 – 7)4
SOLUTION: 4
Apply the Binomial Theorem to expand (a + b) ,
2
where a = 2p and b = −7.
39. (2y + 3)13, 8th term
SOLUTION: First find the coefficient of the 8th term in the
13
expansion of (a + b) , evaluate nCr for n = 13 and r
= 8 – 1 or 7.
41. (4m + 3n)7
SOLUTION: 7
Apply the Binomial Theorem to expand (a + b) ,
where a = 4m and b = 3n.
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Use
to find a power series Page 10
Thus, the binomial coefficient of the 8th term in (a +
b)
13
is 1716. For (2y + 3)
13
to have the form (a + b)
representation of g(x). Indicate the interval on
Use
representation of g(x). Indicate the interval on
which the series converges. Use a graphing
calculator to graph g(x) and the 6th partial sum
of its power series.
42. g(x) =
Study Guide and Review - Chapter 10
Use
to find a power series to find a power series representation of g(x). Indicate the interval on
which the series converges. Use a graphing
calculator to graph g(x) and the 6th partial sum
of its power series.
42. g(x) =
SOLUTION: To find the transformation that relates f (x) to g(x),
use u-substitution. Substitute u for x in f , equate the
two functions, and solve for u.
SOLUTION: To find the transformation that relates f (x) to g(x),
use u-substitution. Substitute u for x in f , equate the
two functions, and solve for u.
Therefore, g(x) = f (5x). for | x | < 1
Replace x with 5x in f (x) =
for | 5x | < 1.
can be represented by the Therefore, g(x) =
Therefore, g(x) = f (5x). for | x | < 1
Replace x with 5x in f (x) =
power series
.
This series converges for | 5x | < 1, which is
for | 5x | < 1.
equivalent to −1 < 5x < 1 or
can be represented by the Therefore, g(x) =
power series
.
The sixth partial sum S 5(x) of this series is
2
.
3
4
5
or 1 + (5x) + (5x) + (5x) + (5x) + (5x) .
This series converges for | 5x | < 1, which is
equivalent to −1 < 5x < 1 or
A power series representation of g(x) is
.
The sixth partial sum S 5(x) of this series is
2
3
;
.
4
5
or 1 + (5x) + (5x) + (5x) + (5x) + (5x) .
A power series representation of g(x) is
;
Use a graphing calculator to graph g(x) and S 5(x)
on the same screen.
.
Use a graphing calculator to graph g(x) and S 5(x)
on the same screen.
43. g(x) =
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SOLUTION: Page 11
To find the transformation that relates f (x) to g(x),
use u-substitution. Substitute u for x in f , equate the
;
.
Study
and Review - Chapter 10
43. g(x)Guide
=
Use a graphing calculator to graph g(x) and S 5(x)
on the same screen.
SOLUTION: To find the transformation that relates f (x) to g(x),
use u-substitution. Substitute u for x in f , equate the
two functions, and solve for u.
Use the 5th partial sum of the exponential
series to approximate each value. Round to
three decimal places.
44. SOLUTION: . Therefore,
Replace x with
.
for | x | < 1 f(x) =
for Therefore, g(x) =
< 1.
45. e–1.5
can be represented by the power series
SOLUTION: .
< 1, which is This series converges for
equivalent to −1 <
< 1 or .
The sixth partial sum S 5(x) of this series is
or 1 + + + + SOLUTION: + .
A power series representation of g(x) is
;
Find the value of each natural logarithm in the
complex number system.
46. ln (–4)
47. ln (–7.15)
SOLUTION: .
Use a graphing calculator to graph g(x) and S 5(x)
on the same screen.
48. RETAIL A chain of retail coffee stores has a
business plan that includes opening 6 new stores
nationwide annually. If there were 480 stores on
January 1, 2012, how many stores will there be at
the end of the year in 2018?
SOLUTION: eSolutions Manual - Powered by Cognero
Write the first few terms of the sequence describing
Page 12
the number of stores there will be after every year.
The chain starts with 480 stores, so a 0 = 480. At the
47. ln (–7.15)
SOLUTION: Study Guide and Review - Chapter 10
48. RETAIL A chain of retail coffee stores has a
business plan that includes opening 6 new stores
nationwide annually. If there were 480 stores on
January 1, 2012, how many stores will there be at
the end of the year in 2018?
SOLUTION: Write the first few terms of the sequence describing
the number of stores there will be after every year.
The chain starts with 480 stores, so a 0 = 480. At the
end of each year there are 6 more stores. So, the
difference between every term is 6. The first four
terms of the sequence are 480, 486, 492, and 498.
Write an equation to represent this sequence. Each
term a n in this sequence can be found by adding
multiple n of 6 to an initial value of 480. So an
explicit formula for this sequence is a n = 480 + 6n.
The question asks for the number of stores at the
end of 2018 or after 7 years. Use the explicit
formula you wrote to find a 7.
a 7 = 480 + 6(7) = 522
The total number of stores by the end of 2018 is 522.
49. DANCE Mara’s dance team is performing at a
recital. In one formation, there are three dancers in
the front row and two additional dancers in each row
behind the first row.
a. How many dancers are in the fourth row?
b. If there are eight rows, how many members does
the dance team have?
SOLUTION: a. The first row has 3 dancers, the second row has 5
dancers, and the third row has 7 dancers, so a 1 = 3,
a 2 = 5, and a 3 = 7. Find the common difference.
5– 3=2
7– 5=2
The common difference is 2. Add 2 to the third term
to find the fourth term.
7+2=9
So, there are 9 dancers in the 4th row.
b. To find the sum of the dancers in all 8 rows, use
the sum of a finite arithmetic series. Substitute a 1 =
3, n = 8, and d = 2 in the formula for the sum of a
finite arithmetic series.
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end of 2018 or after 7 years. Use the explicit
formula you wrote to find a 7.
a 7 = 480 + 6(7) = 522
The total number of stores by the end of 2018 is 522.
49. DANCE Mara’s dance team is performing at a
recital. In one formation, there are three dancers in
the front row and two additional dancers in each row
behind the first row.
a. How many dancers are in the fourth row?
b. If there are eight rows, how many members does
the dance team have?
SOLUTION: a. The first row has 3 dancers, the second row has 5
dancers, and the third row has 7 dancers, so a 1 = 3,
a 2 = 5, and a 3 = 7. Find the common difference.
5– 3=2
7– 5=2
The common difference is 2. Add 2 to the third term
to find the fourth term.
7+2=9
So, there are 9 dancers in the 4th row.
b. To find the sum of the dancers in all 8 rows, use
the sum of a finite arithmetic series. Substitute a 1 =
3, n = 8, and d = 2 in the formula for the sum of a
finite arithmetic series.
So, if there are 8 rows of dancers, the dance team
has 80 members.
50. AGRICULTURE The population of a herd of cows
increases as shown over the course of four years.
a. Write an explicit formula for finding the population
of the herd after n years. Assume that the sequence
shown above is geometric.
b. What will the population of the herd be after 7
years?
c. About how many years will it take the population
of the herd to exceed 85?
SOLUTION: Page 13
The first year there are 47 cows, the second year
there are 51 cows, and the third year there are 56.
shown above is geometric.
b. What will the population of the herd be after 7
years?
Study
Guide
10
c. About
howand
manyReview
years will-itChapter
take the population
of the herd to exceed 85?
SOLUTION: The first year there are 47 cows, the second year
there are 51 cows, and the third year there are 56.
Therefore, a 1 = 47, a 2 = 51, and a 3 = 56. The
common ratio is 1.09.
= 1.
a. Prove that 0.9 + 0.09 + 0.009 + . . . +
= for any positive integer n.
b. Use your understanding of limits and the
statement you proved in part a to explain why
1 is true.
= SOLUTION: a. Let Pn be the statement 0.9 + 0.09 + 0.009 + . . .
= is a For an explicit formula, substitute a 1 = 47 and r =
+
. Because 0.9 =
1.09 in the nth term formula.
true statement, Pn is true for n = 1. Assume that 0.9
+ 0.09 + 0.009 + . . . +
b. Use the explicit formula you found in part a to find
a 7.
= . is true for
a positive integer k. Show that Pk + 1 must be true.
Therefore, after 7 years 78 cows will be in the herd.
This final statement is exactly Pk + 1, so Pk + 1 is
true. Because Pn is true for n = 1 and Pk implies Pk
c. Substitute a n = 85, a 1 = 47, and r = 1.09 into the
equation found in part a.
+ 1, Pn
is true for n = 2, n = 3, and so on. That is, by
the principle of mathematical induction, 0.9 + 0.09 +
= 0.009 + . . . +
is true for all
positive integers n.
b.
= 0.9 + 0.09 + 0.009 + . . . +
, and
= . For larger and
larger values of n, the value of
is a smaller and smaller fraction, approaching 0. Thus, as n goes to
infinity,
Therefore, it will take about 8 years for the
population of the herd to exceed 85 cows.
infinity,
51. NUMBER THEORY Consider the statement
= for any positive integer n.
b. Use your understanding of limits and the
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statement you proved in part to explain why
1 is true.
= 1.
52. BASKETBALL Julie usually makes 4 out of every
= 1.
a. Prove that 0.9 + 0.09 + 0.009 + . . . +
. Therefore, as n approaches
6 free throws that she attempts. What is the
probability that Julie will make 5 out of 6 of the next
free throws that she attempts?
SOLUTION: A success in this situation is Julie making a free
= throw, so p =
and q =
. Each free throw
Page 14
attempt represents a trial, so n = 6. You want to find
smaller fraction, approaching 0. Thus, as n goes to
infinity,
. Therefore, as n approaches
Study Guide and Review - Chapter 10
= 1.
infinity,
52. BASKETBALL Julie usually makes 4 out of every
6 free throws that she attempts. What is the
probability that Julie will make 5 out of 6 of the next
free throws that she attempts?
SOLUTION: A success in this situation is Julie making a free
throw, so p =
and q =
Use the fifth partial sum of the trigonometric series
for cosine to approximate x.
. Each free throw
attempt represents a trial, so n = 6. You want to find
the probability that Julie succeeds 5 times out of
those 6 trials, so let x = 5. To find this probability,
x n –x
find the value of the term nCxp q
in the
n
expansion of (p + q) .
x ≈ ≈ 39.16224
Use the sine ratio to find the length of y.
The probability that Julie will make 5 out of 6 of the
next free throws that she attempts is 26.3%.
Use the fifth partial sum of the trigonometric series
for sine to approximate y.
53. HEIGHT Lina is estimating the height of a tree.
She stands 30 feet from the base and estimates that
her angle of sight to the top of the tree is 40°. If she uses the fifth partial sum of the trigonometric series
for cosine and sine approximated to three decimal
places to calculate the height of the tree, what is
Lina’s estimate?
y ≈ 39.1622 · 0.6427878 ≈ 25.173. Since Lina is looking at the tree from a distance of 6
feet from the ground, the height of the tree is 25.173
+ 6 ≈ 31.173 feet.
SOLUTION: 40° is radians. Use the cosine ratio to find the length of x.
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