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Mathematics Question on Sequence and series

Let SnS_n denote the sum of the cubes of the first nn natural numbers and sns_n denote the sum of the first nn natural numbers. Then ∑r=1nSrsr\sum\limits_{r=1}^{n} \frac{S_{r}}{s_{r}} equals

A

n(n+1)(n+2)6\frac{n\left(n+1\right)\left(n+2\right)}{6}

B

n(n+1)2\frac{n\left(n+1\right)}{2}

C

n2+3n+22\frac{n^{2}+3n+2}{2}

D

None of these

Answer

n(n+1)(n+2)6\frac{n\left(n+1\right)\left(n+2\right)}{6}

Explanation

Solution

∑r=1nSrsr=∑r=1nr2(r+1)24r(r+1)2\sum\limits_{r=1}^{n} \frac{S_{r}}{s_{r}} =\sum\limits _{r=1}^{n} \frac{\frac{r^{2}\left(r+1\right)^{2}}{4}}{\frac{r\left(r+1\right)}{2}} =∑r=1nr(r+1)2=12∑r=1n(r2+r)=\sum\limits _{r=1}^{n} \frac{r\left(r+1\right)}{2} = \frac{1}{2}\sum\limits _{r=1}^{n} \left(r^{2}+r\right) =12[n(n+1)(2n+1)6+n(n+1)2]= \frac{1}{2}\left[\frac{n\left(n+1\right)\left(2n+1\right)}{6} + \frac{n\left(n+1\right)}{2}\right] =n(n+1)(n+2)6= \frac{n\left(n+1\right)\left(n+2\right)}{6}