Find the sum to n terms of the series 52 + 62 + 72 + … + 202

Given series is 5² + 6² + 7² + … + 20²

It’s seen that,

n^th term, a_n = ( n + 4)² = n² + 8n + 16

Then, the sum of n terms of the series can be expressed as

The sum of n terms of a series is given by the equation

\( S_n= \displaystyle\sum_{k=1}^{n}a_k\)

The sum of n terms of the given series is

\( S_n= \displaystyle\sum_{k=1}^{n}(k^2+8k+16)\)

\(\displaystyle\sum_{k=1}^{n}k^2+8 \displaystyle\sum_{k=1}^{n}k+16\)

\( \dfrac{n(n+1)(2n+1)}{6}+ \dfrac{8n(n+1)}{2}+16n\)

Then,

\( (20)^2=(16+4)^2\) is the 16^th term

Therefore,

\(s_{16} = \dfrac{16(16+1)(2\times 16+1)}{6}+ \)

\( \dfrac{8\times 16(16+1)}{2}+16\times 16\)

= \( \dfrac{16(17)(33)}{6}+ \dfrac{(8)(16)(17)}{2}+256\)

= 1496 + 1088 + 256

= 2840