**Question:**

Find the sum of the series $\left\{2^{2}+4^{2}+6^{2}+\ldots .+(2 n)^{2}\right\}$

**Solution:**

We need to find the sum of the series $\left\{2^{2}+4^{2}+6^{2}+\ldots .+(2 n)^{2}\right\}$.

So, we can find it by using summation of the $\mathrm{n}^{\text {th }}$ term of the given series.

The $n^{\text {th }}$ term of the series is $(2 n)^{2}=4 n^{2}$

(Given data)

$a_{n}=4 n^{2}$

Now, sum of the series $S_{n}=\sum_{k=1}^{n} a_{k}$

$S_{n}=\sum_{k=1}^{n} a_{k}=\sum_{k=1}^{n} 4 n^{2}=4 \sum_{k=1}^{n} n^{2}$

Note:

I. Sum of squares of first $n$ natural numbers, $1^{2}+2^{2}+3^{2}+\ldots n^{2}$,

$\sum_{k=1}^{n} k^{2}=\frac{n(n+1)(2 n+1)}{6}$

$S_{n}=4 \sum_{k=1}^{n} n^{2}=4 \frac{n(n+1)(2 n+1)}{6}$

$S_{n}=\left\{22+42+62+\ldots .+(2 n)^{2}\right\}$

$=4 \frac{n(n+1)(2 n+1)}{6}$

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