Question

Find the sum to $n$ terms of the given series whose $n^{th}$ terms is given by ${(2n-1)}^2$

Answer 1

$a_n={(2n-1)}^2={4n}^2-4n+1\therefore S_n={\sum }_{k=1}^n{a}_k={\sum }_{k=1}^n{(4k}^2-4k+1)=4{\sum }_{k=1}^n{k}^2-4{\sum }_{k=1}^{n}k+{\sum }_{k=1}^{n}1=\frac{4n(n+1)(2n+1)}{6}-\frac{4n(n+1)}{2}+n=\frac{2n(n+1)(2n+1)}{3}-2n(n+1)+n=n[\frac{2({2n}^2+3n+1)}{3}-2(n+1)+1]=n\left[\frac{{4n}^2+6n+2-6n-6+3}{3}\right]=n\left[\frac{{4n}^2-1}{3}\right]$