<s>[INST] Discrete mathematics question $$(n+1)^2+(n+2)^2+(n+3)^2+\dots+(2n)^2=\frac{n(2n+1)(7n+1)}{6}$$
Prove the statement using mathematical induction.
 [/INST] Let $s(n)=1^2+2^2+...+n^2=\sum_{j=1}^nj^2.$. You want calculate $s(2n)-s(n)$. We know that $s(n)=\frac{n(n+1)(2n+1)}{6}$, then $s(2n)=\frac{2n(2n+1)(4n+1)}{6}$. It follows that
$$s(2n)-s(n)=\frac{2n+1}{6}(8n^2+2n-n^2-n)=\frac{2n+1}{6}(7n^2+n)=\frac{n(2n+1)(7n+1)}{6}.$$
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