Solve the following

We know that, $S_{r}=1^{3}+2^{3}+3^{3}+\ldots+r^{3}=\left[\frac{r(r+1)}{2}\right]^{2}$

And, $s_{r}=1+2+3+\ldots+r=\frac{r(r+1)}{2}$

As, $\frac{S_{r}}{s_{r}}=\frac{\left[\frac{r(r+1)}{2}\right]^{2}}{\left[\frac{r(r+1)}{2}\right]}=\frac{r(r+1)}{2}=\frac{1}{2}\left(r^{2}+r\right)$

Now,

$\sum_{r=1}^{n} \frac{S_{r}}{s_{r}}=\sum_{r=1}^{n} \frac{1}{2}\left(r^{2}+r\right)$

$=\frac{1}{2}\left(\sum_{r=1}^{n} r^{2}+\sum_{r=1}^{n} r\right)$

$=\frac{1}{2}\left[\frac{n(n+1)(2 n+1)}{6}+\frac{n(n+1)}{2}\right]$

$=\frac{1}{2} \times \frac{n(n+1)}{2} \times\left[\frac{(2 n+1)}{3}+1\right]$

$=\frac{n(n+1)}{4}\left[\frac{2 n+1+3}{3}\right]$

$=\frac{n(n+1)}{4}\left[\frac{2 n+4}{3}\right]$

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

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