Answer
See below.
Work Step by Step
1. Test for $n=1$, we have $1(1+1)(1+2)=6$ is divisible by 6, thus it is true.
2. Assume the statement is true for $n=k$, we have
$k(k+1)(k+2)$ is divisible by 6
3. For $n=k+1$, we have $(k+1)(k+2)(k+3)=k(k+1)(k+2)+3(k+1)(k+2)$ which contains two parts, the first part is divisible by 6 (step 2), the second part contains a 3 and a 2 (on of k+1 or k+2 is even) so it is also divisible by 6, thus it is true for $n=k+1$
4. We conclude that the statement is true for any $n$.
