Proving Induction $(1\cdot2\cdot3)+(2\cdot3\cdot4)+...+k(k+1)(k+2)=k(k+1)(k+2)(k+3)/4$

Question

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!

Answer 1

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.

Answer 2

I'm stuck with the algebra here and not sure how to simply LHS.

To complete your computation of the LHS in your post, fill the dots in the identity below and simplify:$$\frac{k(k+1)(k+2)(k+3)}4+(k+1)(k+2)(k+3)=\frac{(k+1)(k+2)(k+3)}4\,\left(\cdots+\cdots\right)$$ In other words, keep things as factored as possible.