11.1 SOLUTIONS
485
CHAPTER ELEVEN
Solutions for Section 11.1
1.Adding the terms, we see that
2.Adding the terms, we see that
3.
The sum can be rewritten as
2 + 2(2) + 2(2
2
) +
· · ·
+ 2(2
9
)
.
This is a finite geometric series with
a
= 2
,
r
= 2
, and
n
= 10
. We have
Sum
=
2(1

(2)
10
)
1

2
= 2046
.
4.
We use the formula for the sum of a finite geometric series with
a
= 20
,
r
= 1
.
4
, and
n
= 9
. We have
Sum
=
20(1

(1
.
4)
9
)
1

1
.
4
= 983
.
05
.
5.
This is an infinite geometric series with
a
= 1000
and
r
= 1
.
08
. Since
r >
1
, the series diverges and the sum does not
exist.
6.
We use the formula for the sum of a finite geometric series with
a
= 500
,
r
= 0
.
6
, and
n
= 16
. We have
Sum
=
500(1

(0
.
6)
16
)
1

0
.
6
= 1249
.
65
.
7.
This is an infinite geometric series with
a
= 30
and
r
= 0
.
85
. Since

1
< r <
1
, the series converges and we have
Sum
=
30
1

0
.
85
= 200
.
8.
This is an infinite geometric series with
a
= 25
and
r
= 0
.
2
. Since

1
< r <
1
, the series converges and we have
Sum
=
25
1

0
.
2
= 31
.
25
.
486
Chapter Eleven /SOLUTIONS
9.
We use the formula for the sum of a finite geometric series with
a
= 1
,
r
= 1
/
2
, and
n
= 9
. We have
Sum
=
1

(0
.
5)
9
1

0
.
5
= 1
.
9961
.
10.
This is an infinite geometric series with
a
= 1
and
r
= 1
/
3
. Since

1
< r <
1
, the series converges and we have
Sum
=
1
1

1
/
3
= 1
.
5
.
11.
This is a finite geometric series with
a
= 3
,
r
= 1
/
2
, and
n

1 = 10
, so
n
= 11
. Thus
Sum
=
3(1

(1
/
2)
11
)
1

1
/
2
= 3
·
2
(2
11

1)
2
11
=
3(2
11

1)
2
10
.
12.
Each term in this series is
1
.
5
times the preceding term, so this is an infinite geometric series with
a
= 1000
and
r
= 1
.
5
.
Since
r >
1
, the series diverges and the sum does not exist.
13.
Each term in this series is half the preceding term, so this is an infinite geometric series with
a
= 200
and
r
= 0
.
5
. Since

1
< r <
1
, the series converges and we have
Sum
=
200
1

0
.
5
= 400
.
14.
This is an infinite geometric series with
a
=

2
and
r
=

1
/
2
. Since

1
< r <
1
, the series converges and
Sum
=

2
1

(

1
/
2)
=

4
3
.
15.
Since
a
= 10
and
r
= 0
.
75
, we find the partial sums using the formula
S
n
=
10(1

(0
.
75)
n
)
1

0
.
75
.
For
n
= 5
, we have
S
5
=
10(1

(0
.
75)
5
)
1

0
.
75
= 30
.
51
.
For
n
= 10
, we have
S
10
=
10(1

(0
.
75)
10
)
1

0
.
75
= 37
.
75
.
For
n
= 15
, we have
S
15
=
10(1

(0
.
75)
15
)
1

0
.
75
= 39
.
47
.
For
n
= 20
, we have
S
20
=
10(1

(0
.
75)
20
)
1

0
.
75
= 39
.
87
.
As
n
gets larger, the partial sums appear to be approaching
40
, as we expect.
16.
We find the partial sums using the formula
S
n
=
250(1

(1
.
2)
n
)
1

1
.
2
.
For
n
= 5
, we have
S
5
=
250(1

(1
.
2)
5
)
1

1
.
2
= 1
,
860
.
40
.
For
n
= 10
, we have
S
10
=
250(1

(1
.
2)
10
)
1

1
.
2
= 6
,
489
.
67
.
For
n
= 15
, we have
S
15
=
250(1

(1
.
2)
15
)
1

1
.
2
= 18
,
008
.
78
.
For
n
= 20
, we have
S
20
=
250(1

(1
.
2)
20
)
1

1
.
2
= 46
,
672
.
00
.
As
n
gets larger, the partial sums appear to be growing without bound, as we expect, since
r >
1
.
11.1 SOLUTIONS
487
17.
(a)
Notice that the
6
th
deposit is made 5 months after the first deposit, so the first deposit has grown to
500(1
.
005)
5
at
that time. The balance in the account right after the
6
th
deposit is the sum
Balance
= 500 + 500(1
.
005) + 500(1
.
005)
2
+
· · ·
+ 500(1
.
005)
5
.