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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
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The correct answer is:
$m(2 m+1)^2$
$m(2 m+1)^2$
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 $\frac{2 m\left(2 m+1^2\right.}{2} \leqslant m\left(2 m+1^2\right)$
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