Question

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

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Answer 1

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

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

$n^{th}$ 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

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