Find the sum to n terms of the series whose nth terms is given by (2n – 1)2

$a_{n}=(2 n-1)^{2}=4 n^{2}-4 n+1$

$\therefore S_{n}=\sum_{k=1}^{n} a_{k}=\sum_{k=1}^{n}\left(4 k^{2}-4 k+1\right)$

$=4 \sum_{k=1}^{n} k^{2}-4 \sum_{k=1}^{n} k+\sum_{k=1}^{n} 1$

$=\frac{4 n(n+1)(2 n+1)}{6}-\frac{4 n(n+1)}{2}+n$

$=\frac{2 n(n+1)(2 n+1)}{3}-2 n(n+1)+n$

$=n\left[\frac{2\left(2 n^{2}+3 n+1\right)}{3}-2(n+1)+1\right]$

$=n\left[\frac{4 n^{2}+6 n+2-6 n-6+3}{3}\right]$

$=n\left[\frac{4 n^{2}-1}{3}\right]$

$=\frac{n(2 n+1)(2 n-1)}{3}$