Question

The sum of n terms of the series $1+2(1+\frac{1}{n})+3{(1+\frac{1}{n})}^2+.....$ is

• A. $n^2$
• B. $n(n+1)$
• C. $n{(1+\frac{1}{n})}^2$
• D. ${(n+1)}^2$

Answer 1

The correct option is A $n^2$

$S=1+2(1+\frac{1}{n})+3{(1+\frac{1}{n})}^2+.....n{(1+\frac{1}{n})}^{n-1}$

$-S(1+\frac{1}{n})=1(1+\frac{1}{n})+2{(1+\frac{1}{n})}^2+.......+n{(1+\frac{1}{n})}^n$

$S(1-1-\frac{1}{n})=1+(1+\frac{1}{n})+{(1+\frac{1}{n})}^2+......+{(1+\frac{1}{n})}^{n-1}-n{(1+\frac{1}{n})}^n$

$\frac{-S}{n}=\frac{{(1+\frac{1}{n})}^n-1}{(1+\frac{1}{n})-1}-n{(1+\frac{1}{n})}^n$

$\frac{-S}{n}=+n[{(1+\frac{1}{n})}^n-1]-n{(1+\frac{1}{n})}^n$

$\frac{-S}{n}=-n$

$S=n^2$