sum of first nth term of the seris 2^2+4^2+6^2+8^2+.........n terms is
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Answer:
(2/3)n(n+1)(2n+1)
Step-by-step 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)