• # question_answer $\sum\limits_{r=1}^{20}{\,\,\sum\limits_{p=1}^{r}{{{p}^{2}}}}$is equal to A)  16170                   B)  13300                C)  16100                   D)  None of these

We have given that $\sum\limits_{r=1}^{20}{\sum\limits_{p=1}^{r}{{{p}^{2}}}=\sum\limits_{r=1}^{20}{\frac{r(r+1)(2r+1)}{6}}}$ $=\frac{1}{6}\sum\limits_{r=1}^{20}{r(2{{r}^{2}}+3r+1)}$ $=\frac{1}{6}\sum\limits_{r=1}^{20}{(2{{r}^{3}}+3{{r}^{2}}+r)}$ $=\frac{1}{6}\left[ 2\sum\limits_{r=1}^{20}{{{r}^{3}}+3\sum\limits_{r=1}^{20}{{{r}^{2}}+}\sum\limits_{r=1}^{20}{r}} \right]$ $=\frac{1}{6}\left[ 2{{\left( \frac{20(20+1)}{2} \right)}^{2}}+3\frac{20(20+1)(40+1)}{6}+\frac{20(20+1)}{2} \right]$ $=\frac{1}{6}[88200+8610+210]=\frac{97020}{6}=16170$