For any integer $n\geq2$, prove that $\displaystyle\sum\limits\_{i=1}^{n-1}i(i+1) = \frac{n(n-1)(n+1)}{3}.$

---

Let $P(n)$ be the formula $\displaystyle\sum\limits\_{i=1}^{n-1}i(i+1) = \frac{n(n-1)(n+1)}{3}.$

$P(2)$ is true since $\displaystyle\frac{2\cdot1\cdot3}{3}=2.$

$P(k)$ gives $\displaystyle\sum\limits\_{i=1}^{k-1}i(i+1) = \frac{k(k-1)(k+1)}{3}.$

We wish to show that $P(k+1)$ = $\displaystyle\sum\limits\_{i=1}^{k}i(i+1) = \frac{k(k+1)(k+2)}{3}$ is true.

From the left hand side, 

\begin{align}\displaystyle\sum\limits\_{i=1}^{k}i(i+1) &= \displaystyle\sum\limits\_{i=1}^{k-1}i(i+1) + (k+1)((k+1)+1) \\
&=\displaystyle\frac{k(k-1)(k+1)}{3}+(k+1)(k+2) \\
\end{align}

I'm pretty sure I messed up somewhere already, as after all the algebra, I can't have show it of the form $P(k+1)$. How do you these types of inductions? Where is my error? Thanks.