2² + 4² + 6² + 8² + ...

Open in App

Solution

Let $T_{n}$ be the nth term of the given series.

Thus, we have:

$T_{n} = {(2 n)}^{2}$

Now, let $S_{n}$ be the sum of n terms of the given series.

Thus, we have:

$S_{n} = \sum_{k = 1}^{n} T_{k} = \sum_{k = 1}^{n} {(2 k)}^{2} = \sum_{k = 1}^{n} 4 k^{2} = 4 \sum_{k = 1}^{n} k^{2} = 4 \frac{n (n + 1) (2 n + 1)}{6} = \frac{2 n}{3} (n + 1) (2 n + 1)$

Suggest Corrections

Similar questions

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

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

1^{2} + 2^{2} - 3^{2} + 4^{2} + 5^{2} - 6^{2} + 7^{2} + 8^{2} - 9^{2} + . . . .

3 n

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

n

2^{2} + 4^{2} + 6^{2} + 8^{2} + . . .

Join BYJU'S Learning Program