00:01
In this problem, the nth partial sum of a series is given.
00:04
We need to use that to answer a few questions.
00:07
In the first question, we have been asked to find the nth term of the series a -n.
00:12
Now, to obtain the nth term of the series, we need to calculate s -n minus sn -1.
00:19
That is, we need to subtract the partial sum of the first n -1 terms of the series from the partial sum of the first n terms of the series.
00:28
That will give us the nth term a -n of the series.
00:32
So s -n is given to be 3 minus n times 2 -the -power minus n.
00:39
And s -n -1 will be obtained by replacing n -by -n -1 over here, so that we end up with 3 -n -n -minus 1, 2 -to -the -power minus n -1.
00:53
So we have 3 -n -d -n -d -d -d -d -d -d -d -d -d -t divided by 2 -to -the -power n, minus 3, plus, n minus 1 divided by 2 to the power n minus 1.
01:06
So we can cancel these two out, and we end up with n minus 1 by 2 to the power n minus 1, minus n divided by 2 to the power n.
01:16
So the least common denominator over here will be 2 to the power n.
01:21
So 2 to the power n minus 1 times 2 is 2 to the power n.
01:25
So we end up with 2 times n minus 1.
01:29
And over here we have just minus n.
01:32
2n minus 2 minus n divided by 2 to the power n.
01:36
2n minus n is n.
01:38
So we have n minus 2 by 2 to the power n.
01:42
Hence the required n -th term of the series is n -minus 2 divided by 2 -d -power n.
01:49
In the next subpart, we have been asked to determine the sum, summation, n equals to 1 to infinity, a n...