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The sum to $n$ terms of the series $2^2+4^2+6^2+$\ldots\ldotsis
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Verified Answer
The correct answer is:
$\frac{2 n(n+1)(2 n+1)}{3}$
$2^2+4^2+6^2+\ldots \ldots \ldots+(2 n)^2$ $=2^2\left(1^2+2^2+3^2+\ldots \ldots . .+n^2\right)$
$=\frac{4 n(n+1)(2 n+1)}{6}=\frac{2 n(n+1)(2 n+1)}{3}$
$=\frac{4 n(n+1)(2 n+1)}{6}=\frac{2 n(n+1)(2 n+1)}{3}$
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