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 $100S_n=n,$ then $n$ is equal to :

• A. 199
• B. 99
• C. 200
• D. 19

Answer 1

Answer: (A)

$T_n=\frac{\frac{n+(n+1)}{2}}{{\left(\frac{n+(n+1)}{2}\right)}^2}$
$T_n=\frac{2}{n(n+1)}$
$S_n=2\sum _{n=1}^n\left(\frac{1}{n}-\frac{1}{n+1}\right)$
$=2\{\begin{array}{l}1-\frac{1}{2}\\ \frac{1}{2}-\frac{1}{3}\end{array}$

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