\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.
First prove the basis of the induction: If $n=1$, then
$$1*3=^? \frac{(1)(1+1)(2(1)+7)}{6}=\frac{1\cdot 2\cdot 9}{6}=3$$
So efectivelly, $1*3=3$. Suppose by induction hypothesis that
$$1*3+2*4+\dots + k*(k+2)=\frac{k(k+1)(2k+7)}{6}$$
Then we prove it for $n=k+1$:
$$1*3+2*4+\dots+ k(k+2)+(k+1)(k+3)=\frac{k(k+1)(2k+7)}{6}+(k+1)(k+3)$$
$$\frac{k(k+1)(2k+7)}{6}+(k+1)(k+3)=(k+1)\left( \frac{k(2k+7)}{6}+k+4\right)$$
$$(k+1)\left( \frac{k(2k+7)}{6}+k+3\right)=\frac{(k+1)}{6}\left( 2k^2+7k+6k+18\right)$$
$$\frac{(k+1)}{6}\left( 2k^2+7k+6k+18\right)=\frac{(k+1)(k+2)(2k+9)}{6}$$
$$\frac{(k+1)((k+1)+2)(2(k+1)+7)}{6}$$
That is exactly the equation for $k+1$. So it's done.
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?
You have missed some parenthesis.
Write the induction step as: $$ 1\cdot 3 +2\cdot 4+\cdots +n(n+2)+(n+1)(n+3)= $$ $$ =\frac{n(n+1)(2n+7)}{6}+(n+1)(n+3)= $$ $$ =\frac{n(n+1)(2n+7)+6(n+1)(n+3)}{6}=\frac{(n+1)(2n^2+13n+18)}{6}= $$ $$ =\frac{(n+1)(2n^2+4n+9n+18)}{6}=\frac{(n+1)[(2n(n+2)+9(n+2)]}{6}= $$ $$ =\frac{(n+1)(n+2)(2n+9)}{6} $$
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.
