The sum to n terms of the series $2^{2} + 4^{2} + 6^{2} + . . . . . . . . . . i s$

Open in App

Solution

$2^{2} + 4^{2} + 6^{2} + . . . . . . . . . . + (2 n)^{2}$

= $(2)^{2} (1^{2} + 2^{2} + 3^{2} + . . . . . . + n^{2}$ )

= $\frac{4 n (n + 1) (2 n + 1)}{6}$ = $\frac{2 n (n + 1) (2 n + 1)}{3}$ .

Suggest Corrections

Similar questions

The sum of n terms of the series $2^{2} + 4^{2} + 6^{2} + . . . . . . . . . . i s$

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

n

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

n

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

n

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

Join BYJU'S Learning Program