Question

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

Then${\sum }_{k=1}^{n}k{S}_k=?$

• A. $\frac{n(n+2)}{2}$
• B. $\frac{n(n-2)}{2}$
• C. $\frac{n(n+6)}{2}$
• D. $\frac{n(n+3)}{2}$

Answer 1

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

$S_k=1+\frac{1}{k+1}+\frac{1}{{(k+1)}^2}+...+\infty$ $=\frac{1}{1-\frac{1}{k+1}}=\frac{k+1}{k}$
$\therefore k{S}_k=k+1$

$\Rightarrow \sum _{k=1}^{n}k{S}_k=\sum _{k=1}^n(k+1)$ $=2+3+...+(n+1)=\frac{n(n+3)}{2}$