Answer : $\;(d)\;\frac{4}{3}\;p-\frac{2}{3}\;\frac{n}{n+1}$
Explanation : $\;a_{r}=\large\frac{\frac{1}{6}\;r(r+1)(2r+1)}{\large\frac{r^2(r+1)^2}{4}}$
$=\frac{2}{3}\;\large\frac{(2r+1)}{r(r+1)}$
$=\frac{2}{3}\;[\large\frac{1}{r}+\large\frac{1}{r+1}]$
$S=\displaystyle\sum_{r=1}^{n}\;a_{r}=\large\frac{2}{3}\;[(1+\large\frac{1}{2})+(\large\frac{1}{2}+\large\frac{1}{3})+....\;(\large\frac{1}{n}+\large\frac{1}{n+1})]$
$S=\large\frac{2}{3}\;[p+p-1+\large\frac{1}{n+1}]$
$S=\large\frac{4}{3}\;p-\large\frac{2}{3}\large\frac{n}{n+1}\;.$