Question

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

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{know}\mathrm{that},S_r=1^3+2^3+3^3+...+r^3={\left[\frac{r\left(r+1\right)}{2}\right]}^2 \\ \mathrm{And},s_r=1+2+3+...+r=\frac{r\left(r+1\right)}{2} \\ \mathrm{As},\frac{S_r}{s_r}=\frac{{\left[{\displaystyle \frac{r\left(r+1\right)}{2}}\right]}^2}{\left[{\displaystyle \frac{r\left(r+1\right)}{2}}\right]}=\frac{r\left(r+1\right)}{2}=\frac{1}{2}\left(r^2+r\right) \\ \mathrm{Now}, \\ \sum _{r=1}^n\frac{S_r}{s_r}=\sum _{r=1}^n\frac{1}{2}\left(r^2+r\right) \\ =\frac{1}{2}\left(\sum _{r=1}^n{r}^2+\sum _{r=1}^{n}r\right) \\ =\frac{1}{2}\left[\frac{n\left(n+1\right)\left(2n+1\right)}{6}+\frac{n\left(n+1\right)}{2}\right] \\ =\frac{1}{2}\times \frac{n\left(n+1\right)}{2}\times \left[\frac{\left(2n+1\right)}{3}+1\right] \\ =\frac{n\left(n+1\right)}{4}\left[\frac{2n+1+3}{3}\right] \\ =\frac{n\left(n+1\right)}{4}\left[\frac{2n+4}{3}\right] \\ =\frac{n\left(n+1\right)}{4}\times \frac{2\left(n+2\right)}{3} \\ =\frac{n\left(n+1\right)\left(n+2\right)}{6} \end{gathered}$$