Question #ff053

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Answer 1 · 2016-12-01T09:28:41

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

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

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