By induction:
It's true for [tex]n=1[/tex], since [tex]1\cdot2\cdot3[/tex] clearly contains a factor of 3.
Suppose it's true for [tex]n=k[/tex], that [tex]k(k+1)(k+2)[/tex] is divisible by 3. Then
[tex](k+1)(k+2)(k+3)=\dfrac{k(k+1)(k+2)(k+3)}k=\dfrac{3m(k+3)}k[/tex]
where [tex]m[/tex] is an integer. This reduces to
[tex]\dfrac{3m(k+3)}k=3m+9\dfrac mk[/tex]
and both terms are clearly multiples of 3. We know that [tex]\dfrac mk[/tex] is an integer since we had set [tex]m=k(k+1)(k+2)[/tex] previously, which implies [tex]m[/tex] is a multiple of [tex]k[/tex]. So the statement is true.
