The sum of the series $1^2+2.2^2+3^2+2.4^2+5^2+2.6^2+\ldots . .+2(2 m)^2$ is
The sum of the series $1^2+2.2^2+3^2+2.4^2+5^2+2.6^2+\ldots . .+2(2 m)^2$ is
$m(2 m+1)^2$
$m^2(m+2)$
$m^2(2 m+1)$
$m(m+2)^2$
Solution
The sum of the given series $1^2+2.2^2+3^2+$ $2.4^2+5^2+2.6^2+\ldots \ldots . .+2(2 m)^2$ is $\frac{2 m\left(2 m+1^2\right.}{2} \leqslant m\left(2 m+1^2\right)$