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

The correct option is A $199$

$S_n=\frac{1}{1^3}+\frac{1+2}{1^3+2^3}+\frac{1+2+3}{1^3+2^3+3^3}+.....+\frac{1+2+......+n}{1^3+2^3+......+n^3}$

$T_r=\frac{\sum _{k=1}^{r}k}{\sum _{k=1}^r{k}^3}=\frac{\frac{r(r+1)}{2}}{{\left[\frac{r(r+1)}{2}\right]}^2}=\frac{2}{r(r+1)}=\frac{2[r+1-r]}{r(r+1)}=\frac{2}{r}-\frac{2}{r+1}$

$S_n=\sum _{r=1}^n\frac{2}{r}-\frac{2}{r+1}$

$=\frac{2}{1}-\frac{2}{2}+\frac{2}{2}-\frac{2}{3}+\frac{2}{3}-\frac{2}{4}+.......+\frac{2}{n}-\frac{2}{n+1}$

$=2-\frac{2}{n+1}$

$\Rightarrow$$100S_n=n\Rightarrow 100(2-\frac{2}{n+1})=n$

$\Rightarrow$$100(2n+2-2)=n(n+1)$

$\Rightarrow$$200n=n(n+1)$

$\Rightarrow n^2-199n=0$

$\Rightarrow n(n-199)=0$

$\Rightarrow n=0orn=199$

$n$ cannot be $0$

Hence, $n=199$.