Question:
Find the sum to n terms of the series $5^{2}+6^{2}+7^{2}+\ldots+20^{2}$
Solution:
The given series is $5^{2}+6^{2}+7^{2}+\ldots+20^{2}$
$n^{\text {th }}$ term, $a_{n}=(n+4)^{2}=n^{2}+8 n+16$
$\therefore S_{n}=\sum_{k=1}^{n} a_{k}=\sum_{k=1}^{n}\left(k^{2}+8 k+16\right)$
$=\sum_{k=1}^{n} k^{2}+8 \sum_{k=1}^{n} k+\sum_{k=1}^{n} 16$
$=\frac{n(n+1)(2 n+1)}{6}+\frac{8 n(n+1)}{2}+16 n$
$16^{\text {th }}$ term is $(16+4)^{2}=20^{2} 2$
$\therefore S_{16}=\frac{16(16+1)(2 \times 16+1)}{6}+\frac{8 \times 16 \times(16+1)}{2}+16 \times 16$
$=\frac{(16)(17)(33)}{6}+\frac{(8) \times 16 \times(16+1)}{2}+16 \times 16$
$=\frac{(16)(17)(33)}{6}+\frac{(8)(16)(17)}{2}+256$
$=1496+1088+256$
$=2840$
$\therefore 5^{2}+6^{2}+7^{2}+\ldots \ldots+20^{2}=2840$