1
00:00:00,920 --> 00:00:02,480
NARRATOR: Exercise 2.

2
00:00:02,480 --> 00:00:05,360
Find the limiting sum
for the geometric series

3
00:00:05,360 --> 00:00:11,280
3 on 2 plus 9 on 8 plus 27 on 32
plus, etc.

4
00:00:12,680 --> 00:00:15,640
In this geometric series,
the first term is 3 on 2.

5
00:00:15,640 --> 00:00:18,120
So a is equal to 3 on 2.

6
00:00:18,120 --> 00:00:20,960
The common ratio, r,
can be determined

7
00:00:20,960 --> 00:00:23,680
by dividing consecutive terms
in the series.

8
00:00:23,680 --> 00:00:26,560
9 on 8 divided by 3 on 2

9
00:00:26,560 --> 00:00:29,640
is equivalent to 9 and 8
multiplied by two-thirds,

10
00:00:29,640 --> 00:00:32,920
which is 18 on 24 or three-quarters.

11
00:00:32,920 --> 00:00:38,120
Similarly, 27 divided by 32
divided by 9 on 8

12
00:00:38,120 --> 00:00:41,880
is equal to 27 on 32
times eight-ninths,

13
00:00:41,880 --> 00:00:44,000
which is also three-quarters.

14
00:00:44,000 --> 00:00:48,520
And so we establish a common ratio,
r equals three-quarters.

15
00:00:48,520 --> 00:00:51,400
The limiting sum
of a geometric series

16
00:00:51,400 --> 00:00:57,120
is S infinity which is equal to
a divided by 1 minus r.

17
00:00:57,120 --> 00:01:01,320
In this case, a is 3 on 2
and r is three-quarters.

18
00:01:01,320 --> 00:01:06,280
So the limiting sum is 3 on 2
divided by 1 minus three-quarters,

19
00:01:06,280 --> 00:01:08,760
that is 3 on 2 divided by a quarter

20
00:01:08,760 --> 00:01:11,480
or 3 on 2 multiplied by 4,

21
00:01:11,480 --> 00:01:14,400
which is 12 on 2
or 6.

22
00:01:14,400 --> 00:01:16,080
So the limiting sum

23
00:01:16,080 --> 00:01:18,120
for the geometric
series is 6.