Proving Induction $(1\cdot2\cdot3)+(2\cdot3\cdot4)+...+k(k+1)(k+2)=k(k+1)(k+2)(k+3)/4$ I need a little help with the algebra portion of the proof by induction. 
Here's what I have:
Basis Step: $P(1)=1(1+1)(1+2)=6=1(1+1)(1+2)(1+3)/4=6$ - Proven
Induction Step: 
$$(1\cdot2\cdot3)+(2\cdot3\cdot4)+...+k(k+1)(k+2)+(k+1)(k+2)(k+3)=(k+1)(k+2)(k+3)(k+4)/4$$
 $$=k(k+1)(k+2)(k+3)/4+(k+1)(k+2)(k+3)=(k+1)(k+2)(k+3)(k+4)/4$$
I'm stuck with the algebra here and not sure how to simply LHS. Any suggestions, or another set of eyes to to see another solution would be great!

Easier: multiply out the brackets to get $k^3 + 3 k^2 +2k$ and then prove induction or perturbation (this is better!) for each sum. Then combine them back to get the answer. 
