Question

The sum to $n$ terms of the series $1+2(1+\frac{2}{n})+3{(1+\frac{2}{n})}^2+4{(1+\frac{2}{n})}^3+...$ is

• A. $\frac{n^2}{4}$
• B. $\frac{n^2}{4}+n$
• C. $\frac{n^2}{4}-n$
• D. None of these

Answer 1

Answer: (D)

None of these

Let given series be
$S=1+2(1+\frac{2}{n})+3{(1+\frac{2}{n})}^2+4{(1+\frac{2}{n})}^3+...+n{(1+\frac{2}{n})}^{n-1}$ .........(1)
Then, $(1+\frac{2}{n})S=(1+\frac{2}{n})+2{(1+\frac{2}{n})}^2+...+(n-1){(1+\frac{2}{n})}^{n-1}+n{(1+\frac{2}{n})}^n$ ...........(2)

Subtracting the second equation from the first, we get

$-\frac{2}{n}S=1+(1+\frac{2}{n})+{(1+\frac{2}{n})}^2+...+{(1+\frac{2}{n})}^{n-1}-n{(1+\frac{2}{n})}^n$
$=\frac{{(1+\frac{2}{n})}^n-1}{1+\frac{2}{n}-1}-n{(1+\frac{2}{n})}^n=\frac{n}{2}{(1+\frac{2}{n})}^n-\frac{n}{2}-n{(1+\frac{2}{n})}^n$
$=-\frac{n}{2}-\frac{n}{2}{(1+\frac{2}{n})}^n$
$\Rightarrow S=\frac{n^2}{4}+\frac{n^2}{4}{(1+\frac{2}{n})}^n$