<s>[INST] 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?
 [/INST] 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.
</s>