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?
$$
\begin{align}
n(n+1)(n+5)
&amp;=n(n+1)(n+2)+3n(n+1)\\[6pt]
&amp;=6\binom{n+2}{3}+6\binom{n+1}{2}
\end{align}
$$



Binomial Coefficient Basics

If, instead of Pascal's Triangle, we define the binomial coefficients as
$$
\binom{n}{k}=\frac{n(n-1)\cdots(n-k+1)}{k!}
$$
Then we have
$$
\begin{align}
\binom{n-1}{k}+\binom{n-1}{k-1}
&amp;=\binom{n-1}{k-1}\frac{n-k}k+\binom{n-1}{k-1}\\
&amp;=\binom{n-1}{k-1}\frac nk\\
&amp;=\binom{n}{k}
\end{align}
$$
Thus, if $\binom{n-1}{k}\in\mathbb{Z}$ for all $k\in\mathbb{Z}$, then $\binom{n}{k}\in\mathbb{Z}$ for all $k\in\mathbb{Z}$.
You just needed one more step.
$$\begin{align}n(n+1)(n+5) = &amp; (n+1)(n+2)(n+6) \\ = &amp; n^3+9 n^2+20 n+12 \\ = &amp; (n^3+6n^2+5n)+3n^2+15n+12 \\ = &amp; n(n+1)(n+5)+\underbrace{3n(n+5)}_{\star}+12\end{align}$$

$\bigstar$ If $n$ is even, then $3n$ is divisible by 6, otherwise $n$ is odd and $3(n+5)$ is divisible by 6.



Of course you could have done this immediately.


$n(n+1)$ is divisible by $2$.
If neither $n$ nor $n+1$ are divisible by $3$, then $n+2$ and $n+5$ will be.


exactly one of the three factors, $n$, $n+1$, $n+5$ is divisible by $3$.

Therefore $n(n+1)(n+5)$ is divisible by $2$ and $3$, and thus by $6$.