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
.