Inductive proof, algebra step

Question

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 $${n(n+1)(2n+1)\over 6}+(n+1)^2 = {(n+1)(n+2)(2n+3)\over 6}$$

Answer 1

\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}

Answer 2

$${n(n+1)(2n+1)\over 6}+(n+1)^2 = (n+1){(2n^2+n+6n+6)\over 6}=(n+1){(2n^2+7n+6)\over 6}={(n+1)(n+2)(2n+3)\over 6}$$

Answer 3

${n(n+1)(2n+1)\over 6}+(n+1)^2=\frac{(n+1)}{6}(2n^2+7n+6)=\frac{(n+1)(n+2)(2n+3)}{6}$