Question

Let $S_n$ denote the sum of the cubes of the first $n$ natural numbers and $s_n$ denote the sum of the first $n$ natural numbers. Then ${\sum }_{r=1}^n\frac{S_r}{s_r}$ equals

• A. $\frac{n(n+1)}{2}$
• B. None of these
• C. $\frac{n(n+1)(n+2)}{6}$
• D. $\frac{n^2+3n+2}{2}$

Answer 1

The correct option is C $\frac{n(n+1)(n+2)}{6}$

It is given that,

$S_n=1^3+2^3+3^3+....+n^3$

$S_n={\left[\frac{n(n+1)}{2}\right]}^2$ ...$\left(i\right)$

and,

$S_n=1+2+3+....+n$

$S_n=\frac{n(n+1)}{2}$ ...$\left(ii\right)$

From $\left(i\right)$ & $\left(ii\right)$,

$S_n=(s_n{)}^2$

$\Rightarrow \frac{S_n}{s_n}=s_n$

$\Rightarrow {\sum }_{r=1}^n\frac{S_r}{s_r}={\sum }_{r=1}^{n}sr$

$\Rightarrow {\sum }_{r=1}^n\frac{S_r}{s_r}={\sum }_{r=1}^n\frac{r(r+1)}{2}$

(From eq. $\left(ii\right)$)

$\Rightarrow {\sum }_{r=1}^n\frac{S_r}{s_r}=\frac{1}{2}{\sum }_{r=1}^n(r^2+r)$

$\Rightarrow {\sum }_{r=1}^n\frac{S_r}{s_r}=\frac{1}{2}{\sum }_{r=1}^n(r^2+r)$

$\Rightarrow {\sum }_{r=1}^n\frac{S_r}{s_r}=\frac{1}{2}[{\sum }_{r=1}^n{r}^2+{\sum }_{r=1}^{n}r]$

$\Rightarrow {\sum }_{r=1}^n\frac{S_r}{s_r}=\frac{1}{2}[\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}]$

$\Rightarrow {\sum }_{r=1}^n\frac{S_r}{s_r}=\frac{n(n+1)}{4}[\frac{2n+1}{3}+1]$

$=\frac{n(n+1)}{4}\times \frac{2(n+2)}{3}$

$=\frac{n(n+1)(n+2)}{6}$

$\therefore {\sum }_{r=1}^n\frac{S_r}{s_r}=\frac{n(n+1)(n+2)}{6}$

$\therefore$ Correct option is $\left(A\right)$