Question

Prove that for all n in the naturals, n(n+1)(n+2) is divisible by 3

Answer 1

By induction:

It's true for $n=1$, since $1\cdot2\cdot3$ clearly contains a factor of 3.

Suppose it's true for $n=k$, that $k(k+1)(k+2)$ is divisible by 3. Then

$(k+1)(k+2)(k+3)=\dfrac{k(k+1)(k+2)(k+3)}k=\dfrac{3m(k+3)}k$

where $m$ is an integer. This reduces to

$\dfrac{3m(k+3)}k=3m+9\dfrac mk$

and both terms are clearly multiples of 3. We know that $\dfrac mk$ is an integer since we had set $m=k(k+1)(k+2)$ previously, which implies $m$ is a multiple of $k$. So the statement is true.