Question

Find the sum to n terms of the series $1^2+3^2+5^2....$ to n terms.

Answer 1

As per given series, nth term becomes,

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

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

$\therefore S_n={\sum }_{k=1}^n(4k^2-4k+1)$

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

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

$=\frac{4\left[n\right(n+1\left)\right(2n+1\left)\right]-12\left[n\right(n+1\left)\right]+6n}{6}$

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

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