Question

Find sum $1^2+3^2+5^2+...+{(2n-1)}^2$

• A. $\frac{n(2n+1)(2n+1)}{3}$.
• B. $\frac{n(2n-1)(2n+1)}{3}$.
• C. $\frac{2n(2n-1)(2n+1)}{3}$.
• D. None of these

Answer 1

The correct option is B $\frac{n(2n-1)(2n+1)}{3}$.

$1^2+2^2+3^2+4^2+5^2+...+{\left(2n-1\right)}^2$

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

Sum of the given series $=$$\sum _{n=1}^n{a}_n$

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

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

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

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

$=\frac{n}{6}\left[8n^2+12n+4+6-12n-12\right]$

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

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

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