Proof that expression is integer, $\frac{(3n)!}{6^nn!}$

Question

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??

Answer 1

Hint: $3k + 2$ and $3k + 1$ are consecutive integers

Answer 2

Hint:

\begin{align*} \frac{(3n)!}{6^n n!} &= \prod_{i=1}^n \frac{(3i-2)(3i-1)(3i)}{6i} \\ &= \prod_{i=1}^n \frac{(3i-2)(3i-1)}{2} \end{align*} Now, justify why $\frac{(3i-2)(3i-1)}{2}$ is an integer, and then $\frac{(3n)!}{6^n n!}$ is the product of a bunch of integers, and hence is itself an integer.

Answer 3

In case you want induction. From where you stopped, note that if $k$ is odd then $3k+1$ is even; whereas if $k$ is even then $3k+2$ is also even. So the quotient $$ \frac{(3k+2)(3k+1)}{2} $$ is an integer.