Use mathematical induction to prove that (n+2)(n+3)(n+7) is divisible by 6. Use mathematical induction to prove that $q(n)=(n+2)(n+3)(n+7)$ is divisible by $6$.
I have already proved the base case at n=1. I need help on the second part to prove $n=k+1$.
What I did: $(n+2)(n+3)(n+7)=6P$
\begin{align*}
((k+2)+1)&((k+1)+3)((k+1)+7) = (k+3)(k+4)(k+8)\\
= &(k+3)[(k+2)+2][(k+7)+1]\\
= &[(k+3)(k+2)+(2)(k+3)][(k+7)+1]\\
= &(k+2)(k+3)(k+7)+2(k+3)(k+7)+(k+2)(k+3)+2(k+3)\\
= &6P+2k^2+20k+42+k^2+5k+6+2k+6\\
= &6P+3k^2+27k+54\\
= &6p+3(k^2+9k+18)
\end{align*}
I'm not sure what to do, my proof turned out to be divisible by 3 instead of 6. Please let me know how I can move forward with this. Thank you!

Let's look at 3 cases:
1) n is divisible by 3: (n+3) is divisible by 3 and one of (n+2) or (n+3) must be even and, therefore, divisible by 2. Being divisible by both 2 and 3, the expression is divisible by 6.
2) n is not divisible by 3 but (n-1) is: (n+2) is divisible by 3 and one of (n+2) or (n+3) must be even. 
3) n is not divisible by 3 but (n-2) is: (n+7) is divisible by 3 and one of (n+2) or (n+3) must be even.

That covers all possible scenarios for n. Thus, the expression is always divisible by 6.
