Question

Find the sum of the series $2^2$ + $4^2$ + $6^2$ + $8^2$ +................ 20 terms __

Answer 1

Given series $2^2$ + $4^2$ + $6^2$ + $8^2$ +.............

Above given series is sum of the square of first 20 even numbers.

Nth term of the series

$t_n$ = $(2n{)}^2$ = $4n^2$

Sum of the n terms

$S_n={\sum }_{i=1}^n{t}_n=4{\sum }_{i=1}^n{n}^2$

= $\frac{4.n(n+1)(2n+1)}{6}$ : substitute n = 20

= $\frac{4\times 20\times 21\times 41}{6}$ = 11480