Question

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

Answer 1

The given series is $5^2+6^2+7^2+...+{20}^2$
nth term , $a_n={(n+4)}^2=n^2+8n+16$
$\therefore S_n={\sum }_{k=1}^n{a}_k={\sum }_{k=1}^n(k^2+8k+16)={\sum }_{k=1}^n{k}^2+8{\sum }_{k=1}^{n}k+{\sum }_{k=1}^{n}16=\frac{n(n+1)(2n+1)}{6}+\frac{8n(n+1)}{2}+16n$
16th term is $(16+4{)}^2={20}^2$
$\therefore S_{16}=\frac{16(16+1)(2\times 16+1)}{6}+\frac{8\times 16\times (16+1)}{2}+16\times 16=\frac{16\left(17\right)\left(33\right)}{6}+\frac{\left(8\right)\times 16\times (16+1)}{2}+16\times 16=\frac{16\left(17\right)\left(33\right)}{6}+\frac{\left(8\right)\left(16\right)\left(17\right)}{2}+256=1496+1088+256=2840\therefore 5^2+6^2+7^2+...+{20}^2=2840$