Sequence-And-Series Question 659
Question: $ \sum\limits_{k=1}^{n}{k{{(1+1/n)}^{k-1}}=} $
Options:
A) $ n(n-1) $
B) $ n(n+1) $
C) $ n^{2} $
D) $ {{(n+1)}^{2}} $
Show Answer
Answer:
Correct Answer: C
Solution:
[c] $ S_{n}=1+2a+3a^{2}+……+n{a^{n-1}} $ or $ S_{n}=a+2a^{2}+……+(n-1){a^{n-1}}+na^{n} $ where, $ a=1+\frac{1}{n} $
$ \therefore (1-a)S_{n}=1+a+a^{2}+…..+{a^{n-1}}-na^{n} $ $ (1-a)S_{n}=\frac{a^{n}-1}{a-1}-na^{n} $
$ \Rightarrow -\frac{1}{n}S_{n}=\frac{{{( 1+\frac{1}{n} )}^{n}}-1}{\frac{1}{n}}-n{{( 1+\frac{1}{n} )}^{n}}=-n $
$ \Rightarrow ,S_{n}=n^{2} $
BETA
