<s>[INST] Proof that expression is integer, $\frac{(3n)!}{6^nn!}$ Can you help me with this exercises?
Proof that expression is integer,
$$\frac{(3n)!}{6^nn!}$$
I've tried for induction!!
$p(1):\frac{(3)!}{6}=1 $
for $p(k)=\frac{(3k)!}{6^kk!}$
for $p(k+1)=\frac{(3k+3)!}{6^{k+1}(k+1)!}$
where,
$$\frac{(3k+3)(3k+2)(3k+1)(3k)!}{6^k.6(k+1)(k!)},$$For hypothesis:
$$\frac{(3k+2)(3k+1)}{2},$$
How can I follow??
help me??
 [/INST] Let us begin with the basic case, $n=0$. The expression $\frac{(3n)!}{6^nn!}$ is an integer for $n=0$, so the basic case is valid. 
Let us assume that the statement is true for $n$, and that the integer it results in is equal to $k$. Our goal is to prove it true for $n+1$.
We first plug in $n+1$ for the expression.
\begin{align*}
\frac{[3(n+1)]!}{6^{n+1}(n+1)!}
\end{align*}
Using factorial properties and properties of exponents, we can rewrite it like this:
\begin{align*}
\frac{(3n+3)(3n+2)(3n+1)(3n)!}{6\cdot6^n(n+1)n!}
\end{align*}
Using our inductive hypothesis, we can rewrite the expression as:
\begin{align*}
k\cdot\frac{(3n+3)(3n+2)(3n+1)}{6(n+1)} 
\end{align*}
We can use algebraic simplification to completely remove the $3n+3$ out of the numerator along with $n+1$, leaving us with:
\begin{align*}
k\cdot\frac{(3n+2)(3n+1)}{2} 
\end{align*}
Notice that $3n+2$ and $3n+1$ are consecutive integers, and the product of consecutive integers is always even, since one of the two integers must be even. This leads to the claim that
\begin{align*}
\frac{(3n+2)(3n+1)}{2}\in\mathbb{Z}
\end{align*}
This means that the entire expression is an integer, and the expression is an integer for $n+1$, finishing the proof.
</s>