Stuck on Induction Problem I have had more than a day of difficulty with this problem. I have watched many youtube videos on induction and have not been able to solve this problem particularly. I would appreciate helpful hints. Heres the question:

Prove: $$\forall n\gt 1, 1*2+2*3+3*4+...+n(n+1)=\frac{(n(n+1)(n+2))}{3}$$

I have shown my work below. I would appreciate some hints and guidance.
My Attempt:
$$1*2+2*3+3*4+...+n(n+1)=\frac{(n(n+1)(n+2))}{3}$$
Base case: $n=1$
$$1(1+1)=\frac{(1(1+1)(1+2))}{3}$$ $$2=\frac{6}{3}=2$$
The base case works.
Induction Hypothesis: $n=k$
$$1*2+2*3+3*4+...+k(k+1)=\frac{(k(k+1)(k+2))}{3}$$
Assume the Induction Hypothesis is true for $k$.
Then, we show: $$\frac{(k(k+1)(k+2))}{3}+(k+1)=\frac{k^2+6k+3}{3}$$
I get pretty confused trying to apply the inductive hypothesis here because it is difficult to treat the $n(n+1)$ as the $n$ since it involves multiplication. 
Again. I would appreciate hints or even an explantation of this problem, since I have exercised my resources on YouTube and reading the text for well over a day now.  
 A: The inductive step consists in showing that 
$$\frac{n(n+1)(n+2)}3+(n+1)(n+2)=\frac{(n+1)(n+2)(n+3)}3,$$
which shouldn't be too hard.
A: You look great right up through here.
Assume:
$1\cdot2+2\cdot3+3\cdot4+...+k(k+1)=\frac{(k(k+1)(k+2))}{3}$
We must show that:
$1\cdot2+2\cdot3+3\cdot4+...+k(k+1)+(k+1)(k+2)=\frac{(k+1)(k+2)(k+3)}{3}$
$1\cdot2+2\cdot3+3\cdot4+...+k(k+1)+(k+1)(k+2) = \frac{(k(k+1)(k+2))}{3}+(k+1)(k+2)\ $ by the inductive hypothesis.
Rather than multiply everything out, notice that $(k+1)(k+2)$ is a common factor.
$(k+1)(k+2)(\frac{k}{3} + 1)\\
\frac{(k+1)(k+2)(k + 3)}{3}\\
$
A: Inductive Hypothesis:-
Assume $P(k)$ is true
$1\cdot2+2\cdot3+3\cdot4+.....+\sum _{ n=1 }^{ k }{ n(n+1) } =\frac{k(k+1)(k+2)}{3}$
Now show that Inductive Hypothesis is true. Show that $P(k+1)$ is true
$$\sum _{ n=1 }^{ k+1 }{ n(n+1) } =\frac{k+1(k+1+1)(k+1+2)}{3}=\frac{k+1(k+2)(k+3)}{3}$$
Now,
$$\sum _{ n=1 }^{ k+1 }{ n(n+1) }=\sum _{ n=1 }^{ k }{ n(n+1) }++(k+1)(k+1+1)$$
$$=\frac{k(k+1)(k+2)}{3}+(k+1)(k+2)$$
$$=\frac{k\color{red}{(k+1)(k+2)}+3\color{red}{(k+1)(k+2)}}{3}$$
$$=\frac{(k+1)(k+2)}{3}[k+3]$$
$$=\frac{(k+1)(k+2)(k+3)}{3}$$
By the principle of mathematical induction $1\cdot2+2\cdot3+3\cdot4+.....+n(n+1)=\frac{n(n+1)(n+2)}{3}$
