Question

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

Answer 1

let $T_n$ be $n^{th}$ term of the series

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

$\sum T_n=\sum 4n^2-4n+1$

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

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

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

$=n(n+1)\left(\frac{4n+2-6}{3}\right)+n$

$=\frac{n}{3}(n+1)(4n-4)+n$

$=\frac{4}{3}n(n+1)(n-1)+n$

$=\frac{4}{3}n(n^2-1)+n$

$=\frac{n}{3}(4n^2-4+3)$

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