<s>[INST] Prove that $n(n+1)(n+5)$ is a multiple of $6$ I need to prove that $n(n+1)(n+5)$ is divisible by 6. where $n$ is a natural number. I have used the method of induction. But not successful
I got the expression $(k^3+6k^2+5k)+3k^2+15k+12$ when $n=k+1$. 
The term inside the bracket is divisible by 6 since we have assumed that the result is true when $n=k$. If we can show that $3k^2+15k+12$ is also divisible by 6, then we are done. But how to proceed? 
 [/INST] Continuing from your start, let $k^3+6k^2+5k=6p$ for some $p\in\mathbb Z$.
Now, 
$$(k+1)(k+2)(k+6)=(k^3+6k^2+5k)+3k^2+15k+12=6p+3k^2+15k+12=6p+3(k^2+5k+4)=6p+3(k+1)(k+4)$$

Case 1: $k$ is even.
Then, $k+4$ is also even. We can thus write $k+4=2q$ for some $q\in\mathbb Z$.
Hence,
$$6p+3(k+1)(k+4)=6p+3(k+1)(2q)=6p+6q(k+1)=6(p+q(k+1))$$
where $(p+q(k+1))\in\mathbb Z$.
Thus, the statement is true when $k$ is even.

Case 2: $k$ is odd.
Then, $k+1$ is  even. We can thus write $k+1=2q$ for some $q\in\mathbb Z$.
Hence,
$$6p+3(k+1)(k+4)=6p+3(k+4)(2q)=6p+6q(k+4)=6(p+q(k+4))$$
where $(p+q(k+4))\in\mathbb Z$.
Thus, the statement is true when $k$ is odd.
</s>