The sum of n terms of the series

Sum of n terms of series  22 + 42 + 62 + .........

Let nth term of series be denoted by⊤n

i.e ⊤n = (2n)2

then $\sum_{r=1}^{n} T_{r}=\sum_{r=1}^{n}(2 r)^{2}$

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

$=4\left[\frac{n(n+1)(2 n+1)}{6}\right]$

i.e sum of $n$ terms of $2^{2}+4^{2}+6^{2}=\frac{2}{3}[n(n+1)(2 n+1)]$