Question

The sum of n terms of the series $2^2+4^2+6^2+..........is$

Answer 1

$2^2+4^2+6^2+..........+(2n{)}^2$

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

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

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