Question

Let $S_k={lim}_{n\to \infty }{\sum }_{i=0}^n\frac{1}{{(k+1)}^i}$. Then ${\sum }_{k=1}^{n}k{S}_k$ equals

• A. $\frac{n(n+1)}{2}$
• B. $\frac{n(n-1)}{2}$
• C. $\frac{n(n+2)}{2}$
• D. $\frac{n(n+3)}{2}$

Answer 1

The correct option is D $\frac{n(n+3)}{2}$

$S_k={lim}_{n\to \infty }{\sum }_{i=0}^n\frac{1}{{(k+1)}^i}$
$\Rightarrow S_k={lim}_{n\to \infty }\left\{\frac{1-{\left(\frac{1}{k+1}\right)}^{n+1}}{1-\left(\frac{1}{k+1}\right)}\right\}$ ....[ sum of G.P series ]
$\Rightarrow S_k=\left(\frac{k+1}{k}\right){lim}_{n\to \infty }(1-\frac{1}{{(k+1)}^{n+1}})$
$\Rightarrow S_k=\frac{k+1}{k}$
$\therefore$ ${\sum }_{k=1}^{n}k{S}_k={\sum }_{k=1}^n(k+1)=\frac{n(n+1)}{2}+n$
$\Rightarrow {\sum }_{k=1}^{n}k{S}_k=\frac{n(n+3)}{2}$
Ans: D