# Prove by induction $\sum_{i=0}^n i(i+1)(i+2) = (n(n+1)(n+2)(n+3))/4$

Anyone knows how to do this? I'm having trouble after the following step:

Prove by induction that $\sum_{i=0}^n i(i+1)(i+2) = (n(n+1)(n+2)(n+3))/4$

Thanks

$((n(n+1)(n+2)(n+3))/4) + (n+1)(n+2)(n+3)$

I'm not sure how to simplify it after this step.

Thanks

If you notice that $$\frac{(n+1)(n+2)(n+3)(n+4)}4-\frac{n(n+1)(n+2)(n+3)}4=\frac{[(n+4)-n](n+1)(n+2)(n+3)}4=(n+1)(n+2)(n+3)$$ the inductive step should be easy.
This follows the idea which can be used in many similar proofs, namely that $$F(n)=\sum_{i=1}^n f(i) \Leftrightarrow F(n)-F(n-1)=f(n), F(0)=0.$$ See this answer by Bill Dubuque.
(I have used $F(n+1)-F(n)$ above, but this does not change too much, maybe you can try to work out $F(n)-F(n-1)$ yourself.)
In case you are already familiar with binomial coefficients, it might be interesting for you that if you divide this equation by six, you can rewrite it as $$\sum_{j=3}^{n+2}\binom j3=\binom{n+3}4,$$ which is a particular case of hockey-stick identity, see e.g. this question: Induction proof concerning a sum of binomial coefficients: $\sum_{j=m}^n\binom{j}{m}=\binom{n+1}{m+1}$