The sum to n terms of the series 2^2 + 4^2 + | vedclass

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(B) The given series is $2^2 + 4^2 + 6^2 + \dots + (2n)^2$.This can be written as $2^2(1^2 + 2^2 + 3^2 + \dots + n^2)$.We know that the sum of the squ

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

Vedclass · Answer

(B) The given series is $2^2 + 4^2 + 6^2 + \dots + (2n)^2$.This can be written as $2^2(1^2 + 2^2 + 3^2 + \dots + n^2)$.We know that the sum of the squares of the first $n$ natural numbers is given by $\sum_{k=1}^{n} k^2 = \frac{n(n + 1)(2n + 1)}{6}$.Therefore,the sum of the series is $4 	imes \frac{n(n + 1)(2n + 1)}{6}$.Simplifying this,we get $\frac{2n(n + 1)(2n + 1)}{3}$.

The sum to $n$ terms of the series $2^2 + 4^2 + 6^2 + \dots$ is

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