Show by Induction, but stuck on Induction Step (Don't know how to implement N = K+1) Prove by Induction that 
  $$
1\cdot 2\cdot3  +  2\cdot3\cdot4 + \cdots + n \cdot(n+1)\cdot(n+2) = \frac{n  (n+1)(n+2)  (n+3)}{4}
$$


For my basis step, I set $$n=0$$
This leaves me with zero on both LHS and RHS, so I proceed.
Induction Hypothesis consists of plugging in $N = K$ assuming the basis step applies for $$ k \leq n $$

After this, I am stuck on the Induction Step. I try to solve for $n = k+1$, but I don't know what to do.

Due to requests asking me to show what I have tried:
$$
(n+1)(n+2)(n+3) = \frac{(n+1)(n+2)(n+3)(n+4)}{4}
$$
I don't know where to go from here. I don't think it is correct anyways.
Just sum both sides $(n+1)(n+2)(n+3)$ yo your Induction Hypothesis. So you get:

$(1)(2)(3) + ... + n(n+1)(n+2) + (n+1)(n+2)(n+3) = \frac{(n)(n+1)(n+2)}{4}+(n+1)(n+2)(n+3)=\frac{(n)(n+1)(n+2)+4(n+1)(n+2)(n+3)}{4}=\frac{(n+1)(n+2)(n+3)(n+4)}{4}$
By assumption of the induction we obtain:
$$1\cdot2\cdot3+\dots+n(n+1)(n+2)+(n+1)(n+2)(n+3)=$$
$$=\frac{n(n+1)(n+2)(n+3)}{4}+(n+1)(n+2)(n+3)=$$
$$=(n+1)(n+2)(n+3)\left(\frac{n}{4}+1\right)=\frac{(n+1)(n+2)(n+3)(n+4)}{4}.$$
Can you end it now?