Proof of $n(n^2+5)$ is divisible by 6 for all integer $n \ge 1$ by mathematical induction 
Prove the following statement by mathematical induction:
$n(n^2+5)$ is divisible by 6 for all integer $n \ge 1$
My attempt:
Let the given statement be p(n).
(1) $1(1^2+5)$=6 Hence, p(1) is true.
(2) Suppose for all integer $k \ge 1$, p(k) is true.
That is, $k(k^2+5)$ is divisible by 6 
We must show that p(k+1) is true.
$(k+1)((k+1)^2+5)$=$k^3+3k^2+3k+1+5(k+1)$
=$k^3+3k^2+8k+6$
=$k(k^2+5)+3k^2+3k+6$
I'm stuck on this step. I feel I have to show $3k^2+3k+6$ is divisible by 6. But, how can I show $3k^2+3k+6$ is divisible by 6?
 A: If $f(n)=n(n^2+5)$
$f(k+1)-f(k)$
$=(k+1)\{(k+1)^2+5\}-k(k^2+5)=3k^2+3k+1+5=6\cdot\dfrac{k(k+1)}2+6$ which is divisible by $6$ as $k(k+1)$ is even
$\implies6\mid f(k)\iff6\mid f(k+1)$
If induction is not mandatory,
$$n(n^2+5)=\underbrace{(n-1)n(n+1)}_{\text{Product of Three consecutive integers }}+6n$$
A: First, show that this is true for $n=1$:
$1(1^2+5)=6$
Second, assume that this is true for $n$:
$n(n^2+5)=6k$
Third, prove that this is true for $n+1$:
$(n+1)((n+1)^2+5)=$
$\color\red{n(n^2+5)}+3n^2+3n+6=$
$\color\red{6k}+3n^2+3n+6=$
$6\left(k+\frac{n(n+1)}{2}+1\right)$

Since either $n$ or $n+1$ is even, $\frac{n(n+1)}{2}$ is integer.
Please note that the assumption is used only in the part marked red.
A: Hint:
let $$f(n)=n(n^2+5),n\geq1$$ then
$$f(n+1)=(n+1)(n^2+2n+6)=n^3+2n^2+6n+n^2+2n+6=n^3+3n^2+8n+6=$$
$$=n^3+5n+3n^2+3n+6=3(n^2+n)+6+n(n^2+5)=f(n)+6\left(\frac{n(n+1)}{2}+1\right)$$
A: I'm utterly confused.  Three hours before asking this, you asked (and had answered) this: Proof of for all integer $n \ge 2$, $n^3-n$ is divisible by 6 by mathematical induction.
In that one you asked, and got answer for, how to show $6| 3k^2 + 3k$.  In this one you are asking for how to show $6| 3k^2 + 3k + 6$.
How can you know the answer to one but not the answer to the other?
Answer to both: $3|3*h$ for any integer h, so $3|3(k^2 + k)$.  If $k$ is odd so is $k^2$ so $k^2 + k$ is the sum of two odd numbers and is even.  If $k$ is even so is $k^2$ and so is $k^2 + k$.  So $k^2 + k$ is even.  So $2|(k^2 +k)$ so $6|3(k^2 + k)$ so $6|3(k^2 + k) + 6$.
BTW:  $n(n^2 + 5) = n^3 + 5n = (n^3 -1) + 6n$  so one is divisible by 6 if and only if the other one is. 
