<s>[INST] Prove by induction that $(n+1)^2 + (n+2)^2 + ... + (2n)^2 = \frac{n(2n+1)(7n+1)}{6}$ Prove by induction that $$(n+1)^2 + (n+2)^2 + ... + (2n)^2 = \frac{n(2n+1)(7n+1)}{6}.$$
I got up to: 
$n=1$ is true, and assuming $n=k$ prove for $n=k+1$.
Prove...
$$\frac{k(2k+1)(7k+1)+6(2(k+1))^2}{6} = \frac{(k+1)(2k+3)(7k+8)}{6}$$
I keep trying to expand to $6(2k+2)^2$ and factorising but I end up being short on one factor, e.g., I end up with $\frac{(k+1)(2k+3)(7k+6)}{6}$.
 [/INST] Hint :
Be careful, at the step $k+1$ the sum is
$$ (k+2)^2+\cdots+(2k+1)^2+(2k+2)^2 $$
and the sum at the step $k$ is
$$ (k+1)^2+\cdots+(2k-1)^2+(2k)^2 $$
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