A) 16170
B) 13300
C) 16100
D) None of these
Correct Answer: A
Solution :
We have given that \[\sum\limits_{r=1}^{20}{\sum\limits_{p=1}^{r}{{{p}^{2}}}=\sum\limits_{r=1}^{20}{\frac{r(r+1)(2r+1)}{6}}}\] \[=\frac{1}{6}\sum\limits_{r=1}^{20}{r(2{{r}^{2}}+3r+1)}\] \[=\frac{1}{6}\sum\limits_{r=1}^{20}{(2{{r}^{3}}+3{{r}^{2}}+r)}\] \[=\frac{1}{6}\left[ 2\sum\limits_{r=1}^{20}{{{r}^{3}}+3\sum\limits_{r=1}^{20}{{{r}^{2}}+}\sum\limits_{r=1}^{20}{r}} \right]\] \[=\frac{1}{6}\left[ 2{{\left( \frac{20(20+1)}{2} \right)}^{2}}+3\frac{20(20+1)(40+1)}{6}+\frac{20(20+1)}{2} \right]\] \[=\frac{1}{6}[88200+8610+210]=\frac{97020}{6}=16170\]
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