Answer
See below,
Work Step by Step
Proofs using mathematical induction consists of two steps:
1) The base case: here we prove that the statement holds for the first natural number.
2) The inductive step: assume the statement is true for some arbitrary natural number $n$ larger than the first natural number, then we prove that then the statement also holds for $n + 1$.
Hence here: 1) For $n=1: $: $6$ is a factor of $1(1+1)(1+2)=6$
2) Assume for $n=k: 6$ is a factor of $k(k+1)(k+2)=k^3 + 3 k^2 + 2 k$.
Then for $n=k+1$: $(k+1)((k+1)+1)((k+1)+2)=k^3 + 6 k^2 + 11 k + 6=(k^3+3k^2+2k)+(3k^2+9k+6)=k(k+1)(k+2)+3(k+1)(k+2)$.
The first term of the sum of is divisible by $6$ by the inductive hypothesis and the second one is divisible by $3$ because of the factor and also divisible by $2$ because $k+1$ and $k+2$ are adjacent numbers thus one of the must be even. Thus that term is also divisible by $6$, thus their sum is also divisible by $6$.
Thus we proved what we wanted to.