Question

Let $S_{n} = \frac{1}{1^{3}} + \frac{1+2}{1^{3} + 2^{3}} + \frac{1+2+3}{1^{3} + 2^{3} + 3^{3}} + ...... + \frac{1+2+...+n}{1^{3} + 2^{3} +.... +n^{3}} . $ . If $100 \, S_n = n , $ then $n$ is equal to :

• A. 199
• B. 99
• C. 200
• D. 19

Answer 1

Answer: (A)

$T_{n}=\frac{\frac{n+(n+1)}{2}}{\left(\frac{n+(n+1)}{2}\right)^{2}}$
$T_{n}=\frac{2}{n(n+1)}$
$S_{n}=2 \displaystyle\sum_{n=1}^{n}\left(\frac{1}{n}-\frac{1}{n+1}\right)$
$=2\begin{cases}1-\frac{1}{2} \\ \frac{1}{2}-\frac{1}{3}\end{cases}$

$=2\left\{1-\frac{1}{n+1}\right\}$
$S_{n}=\frac{2 n}{n+1}$
$100 \times \frac{2 n}{n+1}=n$
$n+1=200$
$n=199$