Proving by induction that $n(n+1)(n+2)$ is divisible by $6$ for all $n \ge 1$ 
Prove with induction that $n(n+1)(n+2)$ is divisible by $6$ for all $n \ge 1$.

Test $n = 1$:
$$1(1+1)(1+2) = 6 $$
Hypothesis:
$$(\exists k \in \mathbb{N})(n(n+1)(n+2) = 6k)$$
This hypothesis is equivalent to
$$(\exists k \in \mathbb{N})(n^3+3n^2+2n = 6k)$$

Prove for $n+1$:
$$(n+1)(n+2)(n+3)$$
Expand it all and notice:
$$\color{blue}{n^3}+\color{blue}{3n^2}+3n^2+9n+\color{blue}{2n}+6$$
Replace with hypothesis:
$$\color{blue}{6k}+3n^2+9n+6$$
Maybe I can factorize that quadratic:
$$6k+(3n+6)(n+1)$$
If I could take $6$ out as a common factor for the whole expression, I would be done. But I am unsure on how to proceed here. How can I achieve that?

$(3n+6)(n+1)=3(n+2)(n+1)$
One of $n+1,n+2$ must be a multiple of 2(they are consecutive).
You have your factor of 6.
