Question

Show that $1^2+3^2+5^2+...+{(2n-1)}^2=\frac{n(2n-1)(2n+1)}{3}.$

Answer 1

$1^2+3^2+5^2+...+{\left(2n-1\right)}^2$

Here $a_n={\left(2n-1\right)}^2=4n^2-4n+1$

Therefore, $1^2+3^2+5^2+...+{\left(2n-1\right)}^2$$=\sum _{n=1}^n{a}_n$

$=\sum _{n=1}^n\left(4n^2-4n+1\right)$

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

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

$=\frac{n}{6}\left[4\left(2n^2+3n+1\right)-12n-12+6\right]$

$=\frac{n}{6}\left[8n^2-2\right]$

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

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