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Question

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

A

$\frac{n\left(n+1\right)\left(2n+1\right)}{6}$

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B

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

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C

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

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D

none of these

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Solution

## The correct option is B $\frac{2n\left(n+1\right)\left(2n+1\right)}{3}$Explanation for the correct option:Find the required sum.Given: series ${2}^{2}+{4}^{2}+{6}^{2}+....n$terms Here ${t}_{n}={\left(2n\right)}^{2}$ $=4{n}^{2}$Sum, $S=\sum 4{n}^{2}$ $=4\sum {n}^{2}\phantom{\rule{0ex}{0ex}}=4\frac{n\left(n+1\right)\left(2n+1\right)}{6}\phantom{\rule{0ex}{0ex}}=\frac{2n\left(n+1\right)\left(2n+1\right)}{3}$Therefore, the sum of $n$ terms of the series ${2}^{2}+{4}^{2}+{6}^{2}+....$is $\frac{2n\left(n+1\right)\left(2n+1\right)}{3}$.Hence option (B) is the correct answer.

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