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

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
$$
\begin{align}
\binom{n-1}{k}+\binom{n-1}{k-1}
&=\binom{n-1}{k-1}\frac{n-k}k+\binom{n-1}{k-1}\\
&=\binom{n-1}{k-1}\frac nk\\
&=\binom{n}{k}
\end{align}
$$
Thus, if $\binom{n-1}{k}\in\mathbb{Z}$ for all $k\in\mathbb{Z}$, then $\binom{n}{k}\in\mathbb{Z}$ for all $k\in\mathbb{Z}$.
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