460 Chapter Eleven /SOLUTIONS
Chapter 11 Exam Questions
Problems and Solutions for Section 11.1
1. Find the sum, if it exists, of
4 + 42 + 43 + · · · + 410
ANSWER:
This is a finite geometric series with a = 4, r = 4, n = 11. Thus
4(1 − 411 )
S= = 5, 592, 404
1−4
2. Find the sum, if it exists, of
100 + 100(1.03) + 100(1.03)2 + 100(1.03)3 + · · ·
ANSWER:
This is an infinite geometric series with a = 100, r = 1.03. Since r > 1, the series diverges.
3. Find the sum, if it exists, of
500 + 250 + 125 + 62.5 + · · ·
ANSWER:
This is an infinite geometric series with a = 500, r = 12 . Since −1 < r < 1, we get
a 500
S= = = 1000
1−r 1 − 21
4. Find the sum, if it exists, of
1 1 1
1− + − + ···
2 4 8
ANSWER:
This is an infinite geometric series with a = 1, r = − 12 . Since −1 < r < 1, we get
a 1 2
S= = =
1−r 1 − − 12 3
5. Each quarter, $1000 is deposited into an account earning 1.2% interest a quarter, compounded quarterly.
(a) How much is in the account right before the 8th deposit?
(b) How much is in the account right after the 12th deposit?
ANSWER:
(a) The quantity in the account after n deposits is represented by the sum
Bn = 1000 + 1000(1.012) + 1000(1.012)2 + · · · + 1000(1.012)n−1 .
This is a finite geometric series with a = 1000, r = 1.012. Right after the 8 th deposit,
a(1 − r n ) 1000(1 − (1.012)8 )
S7 = = ≈ $8344.12.
1−r 1 − 1.012
Therefore, the amount right before the 8th deposit is
8344.12 − 1000 = $7344.12.
(b) Right after the 12th deposit,
a(1 − r n ) 1000(1 − (1.012)12 )
S12 = = ≈ $12, 824.55.
1−r 1 − 1.012
, 11.1 SOLUTIONS 461
6. Estimate the sum
10 + 10(0.3) + 10(0.3)2 + 10(0.3)3 + · · ·
by calculating the partial sums Sn for n = 1, 5, 10, and 50. Verify your estimate by calculating the infinite sum directly.
ANSWER:
For a finite geometric series with a = 10, r = 0.3,
S1 = 10
10(1 − (0.3)5 )
S5 = = 14.251
1 − 0.3
10(1 − (0.3)10 )
S10 = = 14.2856
1 − 0.3
10(1 − (0.3)50 )
S50 = = 14.2857
1 − 0.3
For an infinite series with a = 10, r = 0.3,
10
S= = 14.2857
1 − 0.3
7. Each week, a patient is given a 40 mg dose of an experimental vaccine, and 25% of the vaccine remains in the body after
one week. Find the quantity in the body
(a) Right after the 4th dose.
(b) After 1 year (52 doses).
ANSWER:
(a) The quantity in the body after n doses is represented by the sum
Qn = 40 + 40(0.25) + 40(0.25)2 + · · · + 40(0.25)n−1
This is a finite geometric series with a = 40, r = 0.25. Thus
40(1 − (0.25)4 )
Q4 = = 53.125mg
1 − 0.25
(b)
40(1 − (0.25)52 )
Q52 = ≈ 53.333mg
1 − 0.25
8. Does the infinite series
4 4 4 4
4 + √ + + 3/2 + 2 + · · ·
3 3 3 3
converge or diverge?
ANSWER:
4 4 4 4
4 + √ + + 3/2 + 2 + · · ·
3 3 3 3
is
a + ax + ax2 + ax3 + · · ·
1
with a = 4 and x = √ .
3
1
Since √ < 1, it converges.
3
9. Find the sum of the first 5 and the sum of the first 10 terms of the series in Exercise 8.
ANSWER:
5
4 1− √1
3
S5 = ≈ 8.857
1− √1
3
10
4 1− √1
3
S10 = ≈ 9.425
1− √1
3
,462 Chapter Eleven /SOLUTIONS
10. Find the sum of the series
15
X 4 n
3
n=5
ANSWER:
15
X 5 6 15
4 n 4 4 4
= + + ··· +
3 3 3 3
n=5
5 2 10
4 4 4 4
= 1+ + + ··· +
3 3 3 3
4 5
4 11
3
1− 3
= 4
≈ 286.68
1− 3
a a a a 1 2 1
11. If the sum of the series S = a + + + + is 3 − 3 , what is a?
3 9 27 81 2 3
ANSWER:
1 1 1 1
S =a 1+ + 2 + 3 + 4
3 3 3 3
a 1− 1 5 1
3 a 1− 35 1 1
Sum = 1
= 2
= 3a 1 − 5
1− 3 3
2 3
1 a
= 3a − 4 . So a = 3.
2 3
2
12. A ball is dropped from a height of 14 feet and bounces. Each bounce is of the height of the bounce before.
3
(a) Find an expression for the height to which the ball rises after it hits the floor for the h th time.
(b) Find the total vertical distance the ball has traveled when it hits the floor for the 4 th time.
ANSWER:
(a) Let hn be the height of the nth bounce after the ball hits the floor for the nth time. Then from Figure 11.1.331,
h0 = height before first bounce = 14 feet
2
h1 = height after first bounce = 14 feet
3
2
2
h2 = height after second bounce = 14 feet
3
2 n
Generalizing gives hn = 14 3
.
6
6
14 6
14( 2 )
3
14( 2 )2
3
···
6
hn ···
? ? ? ?
Figure 11.1.331
, 11.1 SOLUTIONS 463
(b) Total vertical distance when the ball hits the floor for the 4th time is
2 3
2 2 2
14 + 2 14 + 2 14 + 2 14 ≈ 53.41 feet
3 3 3
1
13. A tennis ball is dropped from a height of 40 feet and bounces. Each bounce is the height of the bounce before. A
2
3
superball has a bounce the height of the bounce before, and is dropped from a height of 30 feet. Which ball bounces a
4
greater total vertical distance?
ANSWER:
Tennis ball:
6
6
40 6
40( 12 )
40( 1 )2
2
6
40( 1 )3
2
? ? ? ?
Figure 11.1.332
After the first drop of 40 feet, the total vertical distance is a geometric series with a = 40 and r = 1
2
:
1 1 1
40 + 2 · 40( ) + 2 · 40( )2 + 2 · 40( )3 + · · ·
2 2 2
which converges to 40 + 40/(1 − 12 ) = 120 feet.
Superball:
6
6
30 6
30( 34 )
30( 3 )2
4
6
30( 3 )3
4
? ? ? ?
Figure 11.1.333
After the first drop of 30 feet, the total vertical distance is a geometric series with a = 45 and r = 3
4
:
3 3 3
30 + 2 · 30( ) + 2 · 30( )2 + 2 · 30( )3 + · · ·
4 4 4
which converges to 30 + 45/(1 − 34 ) = 210 feet. So the superball travels more vertical distance than the tennis ball.
Chapter 11 Exam Questions
Problems and Solutions for Section 11.1
1. Find the sum, if it exists, of
4 + 42 + 43 + · · · + 410
ANSWER:
This is a finite geometric series with a = 4, r = 4, n = 11. Thus
4(1 − 411 )
S= = 5, 592, 404
1−4
2. Find the sum, if it exists, of
100 + 100(1.03) + 100(1.03)2 + 100(1.03)3 + · · ·
ANSWER:
This is an infinite geometric series with a = 100, r = 1.03. Since r > 1, the series diverges.
3. Find the sum, if it exists, of
500 + 250 + 125 + 62.5 + · · ·
ANSWER:
This is an infinite geometric series with a = 500, r = 12 . Since −1 < r < 1, we get
a 500
S= = = 1000
1−r 1 − 21
4. Find the sum, if it exists, of
1 1 1
1− + − + ···
2 4 8
ANSWER:
This is an infinite geometric series with a = 1, r = − 12 . Since −1 < r < 1, we get
a 1 2
S= = =
1−r 1 − − 12 3
5. Each quarter, $1000 is deposited into an account earning 1.2% interest a quarter, compounded quarterly.
(a) How much is in the account right before the 8th deposit?
(b) How much is in the account right after the 12th deposit?
ANSWER:
(a) The quantity in the account after n deposits is represented by the sum
Bn = 1000 + 1000(1.012) + 1000(1.012)2 + · · · + 1000(1.012)n−1 .
This is a finite geometric series with a = 1000, r = 1.012. Right after the 8 th deposit,
a(1 − r n ) 1000(1 − (1.012)8 )
S7 = = ≈ $8344.12.
1−r 1 − 1.012
Therefore, the amount right before the 8th deposit is
8344.12 − 1000 = $7344.12.
(b) Right after the 12th deposit,
a(1 − r n ) 1000(1 − (1.012)12 )
S12 = = ≈ $12, 824.55.
1−r 1 − 1.012
, 11.1 SOLUTIONS 461
6. Estimate the sum
10 + 10(0.3) + 10(0.3)2 + 10(0.3)3 + · · ·
by calculating the partial sums Sn for n = 1, 5, 10, and 50. Verify your estimate by calculating the infinite sum directly.
ANSWER:
For a finite geometric series with a = 10, r = 0.3,
S1 = 10
10(1 − (0.3)5 )
S5 = = 14.251
1 − 0.3
10(1 − (0.3)10 )
S10 = = 14.2856
1 − 0.3
10(1 − (0.3)50 )
S50 = = 14.2857
1 − 0.3
For an infinite series with a = 10, r = 0.3,
10
S= = 14.2857
1 − 0.3
7. Each week, a patient is given a 40 mg dose of an experimental vaccine, and 25% of the vaccine remains in the body after
one week. Find the quantity in the body
(a) Right after the 4th dose.
(b) After 1 year (52 doses).
ANSWER:
(a) The quantity in the body after n doses is represented by the sum
Qn = 40 + 40(0.25) + 40(0.25)2 + · · · + 40(0.25)n−1
This is a finite geometric series with a = 40, r = 0.25. Thus
40(1 − (0.25)4 )
Q4 = = 53.125mg
1 − 0.25
(b)
40(1 − (0.25)52 )
Q52 = ≈ 53.333mg
1 − 0.25
8. Does the infinite series
4 4 4 4
4 + √ + + 3/2 + 2 + · · ·
3 3 3 3
converge or diverge?
ANSWER:
4 4 4 4
4 + √ + + 3/2 + 2 + · · ·
3 3 3 3
is
a + ax + ax2 + ax3 + · · ·
1
with a = 4 and x = √ .
3
1
Since √ < 1, it converges.
3
9. Find the sum of the first 5 and the sum of the first 10 terms of the series in Exercise 8.
ANSWER:
5
4 1− √1
3
S5 = ≈ 8.857
1− √1
3
10
4 1− √1
3
S10 = ≈ 9.425
1− √1
3
,462 Chapter Eleven /SOLUTIONS
10. Find the sum of the series
15
X 4 n
3
n=5
ANSWER:
15
X 5 6 15
4 n 4 4 4
= + + ··· +
3 3 3 3
n=5
5 2 10
4 4 4 4
= 1+ + + ··· +
3 3 3 3
4 5
4 11
3
1− 3
= 4
≈ 286.68
1− 3
a a a a 1 2 1
11. If the sum of the series S = a + + + + is 3 − 3 , what is a?
3 9 27 81 2 3
ANSWER:
1 1 1 1
S =a 1+ + 2 + 3 + 4
3 3 3 3
a 1− 1 5 1
3 a 1− 35 1 1
Sum = 1
= 2
= 3a 1 − 5
1− 3 3
2 3
1 a
= 3a − 4 . So a = 3.
2 3
2
12. A ball is dropped from a height of 14 feet and bounces. Each bounce is of the height of the bounce before.
3
(a) Find an expression for the height to which the ball rises after it hits the floor for the h th time.
(b) Find the total vertical distance the ball has traveled when it hits the floor for the 4 th time.
ANSWER:
(a) Let hn be the height of the nth bounce after the ball hits the floor for the nth time. Then from Figure 11.1.331,
h0 = height before first bounce = 14 feet
2
h1 = height after first bounce = 14 feet
3
2
2
h2 = height after second bounce = 14 feet
3
2 n
Generalizing gives hn = 14 3
.
6
6
14 6
14( 2 )
3
14( 2 )2
3
···
6
hn ···
? ? ? ?
Figure 11.1.331
, 11.1 SOLUTIONS 463
(b) Total vertical distance when the ball hits the floor for the 4th time is
2 3
2 2 2
14 + 2 14 + 2 14 + 2 14 ≈ 53.41 feet
3 3 3
1
13. A tennis ball is dropped from a height of 40 feet and bounces. Each bounce is the height of the bounce before. A
2
3
superball has a bounce the height of the bounce before, and is dropped from a height of 30 feet. Which ball bounces a
4
greater total vertical distance?
ANSWER:
Tennis ball:
6
6
40 6
40( 12 )
40( 1 )2
2
6
40( 1 )3
2
? ? ? ?
Figure 11.1.332
After the first drop of 40 feet, the total vertical distance is a geometric series with a = 40 and r = 1
2
:
1 1 1
40 + 2 · 40( ) + 2 · 40( )2 + 2 · 40( )3 + · · ·
2 2 2
which converges to 40 + 40/(1 − 12 ) = 120 feet.
Superball:
6
6
30 6
30( 34 )
30( 3 )2
4
6
30( 3 )3
4
? ? ? ?
Figure 11.1.333
After the first drop of 30 feet, the total vertical distance is a geometric series with a = 45 and r = 3
4
:
3 3 3
30 + 2 · 30( ) + 2 · 30( )2 + 2 · 30( )3 + · · ·
4 4 4
which converges to 30 + 45/(1 − 34 ) = 210 feet. So the superball travels more vertical distance than the tennis ball.
