Question

Find the sum of the series 5²+6²+7²+...+20²?​

Answer 1

${5}^{2} + {6}^{2} + {7}^{2} + ... + {20}^{2}$

By Using,

$\displaystyle \sum\limits_{k=1}^{n} k^2 = \dfrac{n(n + 1)(2n + 1)}{6}$

We get,

$\displaystyle \sum\limits_{k=5}^{20} k^2 = \displaystyle \sum\limits_{k=1}^{20} k^2 - \displaystyle \sum\limits_{k=1}^{4} k^2$

$\displaystyle \sum\limits_{k=1}^{n} k^2 = \dfrac{20 \cdot 21 \cdot 41}{6} - \dfrac{4 \cdot 5 \cdot 9 }{6}$

$\displaystyle \sum\limits_{k=5}^{20} k^2 = 2870 - 30 = \boxed{2840}$