Find the sum of the series $2^{2}$ + $4^{2}$ + $6^{2}$ + $8^{2}$ +................ 20 terms ___

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Solution

Given series: $2^{2}$ + $4^{2}$ + $6^{2}$ + $8^{2}$ +.............
Nth term of the series
$t_{n} = (2 n)^{2} = 4 n^{2}$
Sum of the $n$ terms
$S_{n} = \sum_{i = 1}^{n} t_{n} = 4 \sum_{i = 1}^{n} n^{2}$

$= \frac{4 n (n + 1) (2 n + 1)}{6}$
On substituting $n = 20$ , we get-
$= \frac{4 \times 20 \times 21 \times 41}{6} = 11480$

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