<s>[INST] Inductive proof, algebra step I have to prove by induction that 
$$1^2+2^2+3^2+ \cdot \cdot \cdot + n^2 = {n(n+1)(2n+1)\over 6}$$
Base step: $$1^2 = {1(1+1)(2\bullet1 +1)\over 6}$$
$$1^2= {6\over 6} = 1$$
Then I use this  $$1^2+2^2+3^2+ \cdot \cdot \cdot + n^2 +(n+1)^2 = {n(n+1)(2n+1)\over 6}+(n+1)^2$$
to show that
 $$1^2+2^2+3^2+ \cdot \cdot \cdot + n^2 + (n+1)^2 = {(n+1)((n+1)+1)(2(n+1)+1)\over 6}$$
$$=$$
$$1^2+2^2+3^2+ \cdot \cdot \cdot + n^2 + (n+1)^2 = {(n+1)(n+2)(2n+3)\over6}$$
I'm stuck with the algebra part, could someone go through the steps of how 
$${n(n+1)(2n+1)\over 6}+(n+1)^2 = {(n+1)(n+2)(2n+3)\over 6}$$
 [/INST] \begin{align}\frac{n(n+1)(2n+1)}{6}+(n+1)^2 &= \frac{n(n+1)(2n+1)}{6}+\frac{6(n+1)^2}{6}\\
&=\frac{1}{6}\left[(n+1)[n(2n+1)+6(n+1)]\right]\\
&=\frac{1}{6}(n+1)(2n^2+7n+6)\\
&=\frac{1}{6}(n+1)(2n^2+4n+3n+6)\\
&=\frac{1}{6}(n+1)(2n(n+2)+3(n+2))\\
&=\frac{1}{6}(n+1)(n+2)(2n+3)\\
&=\frac{(n+1)((n+1)+1)(2(n+1)+1)}{6}\\
\end{align}
</s>