Solve the following
Question:

22 + 42 + 62 + 82 + …

Solution:

Let $T_{n}$ be the $n$th term of the given series.

Thus, we have:

$T_{n}=(2 n)^{2}$

Now, let $S_{n}$ be the sum of $n$ terms of the given series.

Thus, we have:

$S_{n}=\sum_{k=1}^{n} T_{k}$

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

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

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

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

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