Question

Find the sum of $1^2+3^2+5^2+7^2+\cdots$ to $n$ terms.

Answer 1

Sum of the square of $n$ natural number will be

$1^2+2^2+3^2+4^2+...+n^2=\frac{n(n+1)(2n+1)}{6}$

Thus sum of square of $2n$ natural number wil be

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

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

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

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

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

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

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