# Sum of $n$ terms of series $S$ = $1$ + $2 \;(1+\frac{1}{n})$ + $3(1+\frac{1}{n})^2$ + ....is given by

$(a)\;(n+1)^2\qquad(b)\;n(n+1)\qquad(c)\;n^2\qquad(d)\;\frac{n(n+1)}{2}$

Answer : (c) $n^2$
Explanation : Let $\;1+\frac{1}{n}=x$
$S=1+2x+3x^2+4x^3+....$
$x\;S=x+2x^2+3x^3+4x^4+...$
Subtracting ,
$(1-x)\;S=1+x+x^2+...+x^{n-1}-nx^n$
$=\frac{1-x^n}{1-x}-nx^n$
$\frac{-1}{n}\;S=(-n)\;[1-(1+\frac{1}{n})^n]-n[1+\frac{1}{n}]^n$
$=-n$
$S=n^2.$
answered Jan 20, 2014 by