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!
 A: Hint: You are done, because $k^2+9k+18$ is always even, so that you have divisibility by $2\cdot 3=6$.
A: This is fine so far. You have divisibility by $3$ and only need to check that $k^2+9k+18$ is even to get divisibility by $6$. This is true because $k^2$ and $k$ are either both even or both odd.
A: Easier: call it $f_n.\,$ Check $\,f_n-f_{n-2} = 6(n^2\!+\!8n\!+\!14) =: 6 b_n.$ Base cases $f_0,f_1 = 6(7),6(16)$. Assume for complete induction $\,f_n = 6 a_n\,$ for all naturals below $n\ge 2$. Since $\,f_n = \color{#0a0}{f_{n-2}}+6b_n\,$ and  induction hypothesis $\,\Rightarrow\, \color{#0a0}{f_{n-2}}=6a_{n-2},\:$  we conclude with the sought: $\ f_n = \color{#c00}6a_{n-2}+\color{#c00}6b_n$.
Remark $ $ It's instructive to explain why $\,f_n-f_{n-2}\,$ has the sought factor of $2$ too. By the Factor Theorem if $\,f(n)\,$ is a polyomial in $\,n\,$ with integers coef's then  $\,f(n)-f(n-2)\,$ is divisible by $\,n-(n-2) = 2,\,$ i.e. $\,f(n)-f(n\!-\!2) = 2g(n)\,$ for a polynomial $\,g(n)\,$ with integer coefs. The  factor of $3$ you found in $\,f_n - f_{n-1}\,$ also remains in $\,f_n - f_{n-2} = (f_n-f_{n-1}) + (f_{n-1}-f_{n-2})$ since both braced terms have a factor of $3$. Hence $\,f_n-f_{n-2}\,$ has a factor of $\,{\rm lcm}(2,3) = 6$.
A: Hint:
$k^2+9k+18=(k+3)(k+6)$
Now if k is odd then $k+3$ is even.
If k is even then $k+6$is even. So $k^2+9k+18$is always even.
A: 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.
