Question

Question #ff053

Answer 1

`1/3 n ( 4 n^2-1)`

Explanation:

We know that

`S_(2n)=sum_(k=1)^n(2k)^2+sum_(k=0)^(n-1)(2k+1)^2=sum_(k=1)^(2n)k^2`

then

`sum_(k=0)^(n-1)(2k+1)^2=S_(2n)-4 S_n`

Here

`S_n=n^3/3+n^2/2+n/6`

so

`S_(2n)-4 S_n=1/3 n ( 4 n^2-1)`