Question

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

Answer 1

Given

$a_n=(2n-1{)}^2$

$=4n^2-4n+1$

Sum of n terms is

$S_n={\sum }_{n=1}^n{a}_n$

$={\sum }_{n=1}^{n}4n^2-4n+1$

$={\sum }_{n=1}^{n}4n^2-{\sum }_{n=1}^{n}4n+{\sum }_{n=1}^{n}1$

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

$=2\left(\frac{2(n+1)(2n+1)-6(n+1)+3}{3}\right)$

$=\frac{n}{3}\left[\right(2n+2\left)\right(2n+1)-6n-6+3$

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

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

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

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