<s>[INST] 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.
 [/INST] You still need to prove the base case
Suppose:
$\begin{align}
1*3+2*4+3*5+...+n(n+2) = \frac{n(n+1)(2n+7)}{6}
\end{align}$
show that:
$\begin{align}
1*3+2*4+3*5+...+(n+1)(n+3) = \frac{n+1(n+2)(2n+9)}{6}
\end{align}$
$\begin{align}
1*3+2*4+3*5+...+n+1(n+3) = \frac{n(n+1)(2n+7)}{6} + (n+1)(n+3)
\end{align}$
By the inductive hypothesis.
Can you take it home from here?
</s>