Question

Sum of the terms of the series $1+2(1+\frac{1}{n})+3(1+\frac{1}{n}{)}^2+....$ is given by ________.

• A.
• B.
• C.
• D.

Answer 1

The correct option is B

Let x=1+$\frac{1}{n}$
Given series is an AGP.
$s_n=1+2x+3x^2+......n{x}^{n-1}$ -----(1)
Multiply x in equation 1
$x{s}_n=x+2x^2+....(n-1)x^{n-1}+n{x}^n$ -----(2)
Subtract equation 2 from equation 1, we get,
$(1-x)s_n=1+x+x^2+.....x^{n-1}-n{x}^n$
$(1-x)s_n=\frac{1-x^n}{1-x}-n{x}^n$
$s_n=\frac{(1-x^n)}{(1-x{)}^2}-\frac{n\left(x^n\right)}{(1-x)}$ -----(3)
Since $x=1+\frac{1}{n}$
Substitute $1-x=-\frac{1}{n}$ in equation 3
$s_n=\frac{1-(1+\frac{1}{n}{)}^n}{\left(\frac{-1}{n^2}\right)}-\frac{n(1+\frac{1}{n}{)}^n}{\left(\frac{-1}{n}\right)}$
$=\frac{1-(1+\frac{1}{n}{)}^n+n(1+\frac{1}{n}{)}^2}{\frac{1}{n^2}}$
$=\frac{1}{\frac{1}{n^2}}$
$s_n=n^2$