Question

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

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

Answer 1

Answer: (D)

$ \displaystyle\sum_{r=1}^n \frac{S_r}{S_r} =\sum_{r=1}^n \frac{\frac{r^2(r+1)^2}{4}}{\frac{r(r+1)}{2}} $
$ \left\{\Sigma n^3=\left[\frac{n(n+1)}{2}\right]^2 \text { and sum of the } \text { first } n \text { natural numbers, } \Sigma n=\frac{n(n+1)}{2}\right\}$
$ =\displaystyle\sum_{r=1}^n \frac{r(r+1)}{2}=\frac{1}{2} \displaystyle\sum_{r=1}^n\left(r^2+r\right)$
$ =\frac{1}{2}\left[\displaystyle\sum_{r=1}^n r^2+\sum_{r=1}^n r\right]$
$ =\frac{1}{2}\left[\frac{n(n+1)(2 n+1)}{6}+\frac{n(n+1)}{2}\right]$
$ =\frac{1}{12}(n)(n+1)(2 n+1+3) $
$ =\frac{n(n+1) 2(n+2)}{12}=\frac{n(n+1)(n+2)}{6}$