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

11480

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11875

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12480

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21470

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Solution

The correct option is A 11480
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

$t_{n}$ = $(2 n)^{2}$ = $4 n^{2}$
Sum of the n terms

$S_{n} = \sum_{i = 1}^{n} t_{n} = 4 \sum_{i = 1}^{n} n^{2}$

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

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

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

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