<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] When $n=1$, $LHS=4=RHS.$
Assume, when $n=k$, $LHS=RHS$. When $n=k+1$, 
$LHS=(k+2)^2+(k+3)^2+\dots+(2k+1)^2 + (2(k+1))^2
\\=k(2k+1)(7k+1)/6-(k+1)^2+(2k+1)^2+(2(k+1))^2
\\=[k(2k+1)(7k+1)+18(k+1)^2+6(2k+1)^2]/6
\\=[(2k+1)(7k^2+13k+6)+18(k+1)^2]/6
\\=[(2k+1)(7k+6)(k+1)+18(k+1)^2]/6
\\=(k+1)(14k^2+37k+24)/6
\\=(k+1)(2(k+1)+1)(7(k+1)+1)/6=RHS.$
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