Question

Sum of first nth term of the seris 2^2+4^2+6^2+8^2+.........n terms is​

Answer 1

(2/3)n(n+1)(2n+1)

Explanation:

First you have to find the general term formula which is here (2k)^2 which is equal to 4k^2. Sum(k=1 to n) of 4k^2=4•n(n+1)(2n+1)/6

simplified to (2/3)n(n+1)(2n+1)