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.

You have missed some parenthesis.
Write the induction step as:
$$
1\cdot 3 +2\cdot 4+\cdots +n(n+2)+(n+1)(n+3)=
$$
$$
=\frac{n(n+1)(2n+7)}{6}+(n+1)(n+3)=
$$
$$
=\frac{n(n+1)(2n+7)+6(n+1)(n+3)}{6}=\frac{(n+1)(2n^2+13n+18)}{6}=
$$
$$
=\frac{(n+1)(2n^2+4n+9n+18)}{6}=\frac{(n+1)[(2n(n+2)+9(n+2)]}{6}=
$$
$$
=\frac{(n+1)(n+2)(2n+9)}{6}
$$
