How do I prove this using mathematical induction? \begin{align}
1\cdot3+2\cdot4+3\cdot5+...+n(n+2) = \frac{n(n+1)(2n+7)}{6}
\end{align}
Using the mathematical induction step I arrive at this :
\begin{align}
1\cdot3+2\cdot4+3\cdot5+...+n+1(n+3) = \frac{n+1(n+2)(2n+9)}{6}
\end{align}
And I don't see any other way to continue except to divide
\begin{align}
n+1(n+3)
\end{align}
into
\begin{align}
n(n+2)+something
\end{align}
and substitute it with the beginning of the fraction.
But that doesn't get me anywhere.

All you need to do is realize that
$$\frac{(n+1)(n+2)(2n+9)}6=\frac{(n+1)(n+1+1)(2(n+1)+7)}6$$
which is what you are trying to prove, but with $(n+1)$ everywhere $n$ originally was.
