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 $\frac{(k+1)(2k+3)(7k+6)}{6}$.
 A: Alternative route (also with induction)
You can start proving inductively that: $$\sum_{k=1}^{n}k^{2}=\frac{1}{6}n\left(n+1\right)\left(2n+1\right)$$
Then have a look at: $$\sum_{k=n+1}^{2n}k^{2}=\sum_{k=1}^{2n}k^{2}-\sum_{k=1}^{n}k^{2}$$
A: 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
$$ (k+1)^2+\cdots+(2k-1)^2+(2k)^2 $$
A: Check this out
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$(k+1)^2 +(k+2)^2+ ............ (k+k)^2 = \frac{k(2k+1)(7k+1)}{6}$
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$p(k+1): (k+1)^2+(k+2)^2...........(2k)^2+(2k+1)^2+.(2k+2)^2-(k+1)^2= \frac{(k+1)(2k+3)(7k+8)}{6}$
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$p(k+1):  \frac{k(2k+1)(7k+1)}{6}+3(k+1)^2+(2k+1)^2$=RHS
A: Inductive step spelled out in detail:
\begin{align*}
&\hphantom{=}\frac{n(2n+1)(7n+1)}6+(2n+1)^2+(2n+2)^2-(n+1)^2\\
&=\frac{n(2n+1)(7n+1)}6+(2n+1)^2+3(n+1)^2\\
&=(2n+1)\left(\frac{n(7n+1)}6+2n+1\right)+3(n+1)^2\\
&=(2n+1)\frac{n(7n+1)+12n+6}6+3(n+1)^2\\
&=(2n+1)\frac{7n^2+13n+6 }6+3(n+1)^2\\
&=\frac{(2n+1)(n+1)(7n+6)}6+3(n+1)^2\\
&=(n+1)\left(\frac{(2n+1)(7n+6)}6+3(n+1)\right)\\
&=(n+1)\frac{(2n+1)(7n+6)+18(n+1)}6\\
&=(n+1)\frac{14n^2+19n+6+18n+18}6\\
&=(n+1)\frac{14n^2+37n+24}6\\
&=\frac{(n+1)(2n+3)(7n+8)}6
\end{align*}
