The sum of n terms of the series 22+42+62+....is
n(n+1)(2n+1)6
2n(n+1)(2n+1)3
n(n+1)(2n+1)3
none of these
Explanation for the correct option:
Find the required sum.
Given: series 22+42+62+....nterms
Here tn=(2n)2
=4n2
Sum, S=∑4n2
=4∑n2=4n(n+1)(2n+1)6=2n(n+1)(2n+1)3
Therefore, the sum of n terms of the series 22+42+62+....is 2n(n+1)(2n+1)3.
Hence option (B) is the correct answer.
The sum of n terms of the series 22+42+62+..........is
The sum to n terms of the series 22+42+62+..........is