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hi guys can you help me with this?

Proof that expression is integer

$$\frac{(2n)!}{2^nn!}$$

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marked as duplicate by user940, user147263, colormegone, Leucippus, davidlowryduda Aug 23 '15 at 4:21

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\begin{align} \frac{(2n)!}{2^n n!} = \frac{(2n) \times (2n - 1) \times \cdots \times 1}{(2n) \times (2n - 2) \times \cdots \times 2} = (2n - 1) \times (2n - 3) \times \cdots \times 1 \in \mathbb{Z}. \end{align}

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If $n=1$ then $(2n)!/2^{n}n! = 1$. Let $n \geq 1$. If $(2n)!/2^{n}n! = k$ for some $k \in \mathbb{N}$, then $$ \frac{(2n+2)!}{2^{n+1}(n+1)!} = \frac{(2n+2)(2n+1)(2n)!}{(2n+2)2^{n}n!} = (2n+1)k \in \mathbb{N}, $$ qed.

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