Prove $3|n(n+1)(n+2)$ by induction I tried proving inductively but I didn't really go anywhere. So I tried:

Let $3|n(n+1)(n+2)$.
Then $3|n^3 + 3n^2 + 2n \Longrightarrow  3|(n(n(n+3)) + 2)$

But then?

Consider the binomial $(x+1)^{n+2}$.  The coefficient of the $x^3$ term is
$${n+2\choose 3}={(n+2)!\over 3!(n-1)!}={n(n+1)(n+2)\over 6}$$
Every coefficient of $(x+1)^n$ is an integer for $n$ an integer, therefore $6|n(n+1)(n+2)$ and thus $3|n(n+1)(n+2)$.
Note that this mechanism can apply to any integer, including showing that $1000|n(n+1)(n+2)\cdots(n+998)(n+999)$.
As noted elsewhere, this property of the binomial coefficients is provable by induction, which demonstrates that an inductive proof is truly the proper way to show the question's property.  In binomial terms, the inductive step occurs by noting that
$${n\choose k-1}+{n\choose k}={n+1\choose k}$$
