Question

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

• A. $\frac{n(n+1)(2n+1)}{6}$
• B. $\frac{2n(n+1)(2n+1)}{3}$
• C. $\frac{n(n+1)(2n+1)}{3}$
• D. none of these

Answer 1

The correct option is B

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

Explanation for the correct option:

Find the required sum.

Given: series $2^2+4^2+6^2+....n$terms

Here $t_n={\left(2n\right)}^2$

$=4n^2$

Sum, $S=\sum 4n^2$

$$\begin{gathered} =4\sum n^2 \\ =4\frac{n(n+1)(2n+1)}{6} \\ =\frac{2n(n+1)(2n+1)}{3} \end{gathered}$$

Therefore, the sum of $n$ terms of the series $2^2+4^2+6^2+....$is $\frac{2n(n+1)(2n+1)}{3}$.

Hence option (B) is the correct answer.