Show that for $n\geqslant1,\ \displaystyle \binom{2n}{n} = \dfrac{1\cdot3\cdot5\cdot\cdots\cdot(2n-1)}{2\cdot4\cdot6\cdot\cdots\cdot2n}2^{2n}$

Question

Show that for $n\geqslant1,$ $$\displaystyle \binom{2n}{n} = \dfrac{1\cdot3\cdot5\cdot\cdots\cdot(2n-1)}{2\cdot4\cdot6\cdot\cdots\cdot2n}2^{2n}.$$

This is an exercise from the Preliminaries of David Burton's Elementary Number Theory.

I started by simplifying the $\text{RHS}$ into something that could enable me to use mathematical induction.

$$ \begin{align*} \text{RHS} &= \dfrac{\displaystyle\sum_{k=1}^{n}(2k-1)}{\displaystyle\sum_{k=1}^{n}2k}\cdot2^{2n}\\ &=\frac{n(n+1)-n}{n(n+1)}\cdot2^{2n}\\ &=\frac{2^{2n}n}{n+1}. \end{align*} $$ To be proved is that $$\binom{2n}{n}=\frac{2^{2n}n}{n+1}.$$ To get $\binom{2n}{n}$ on the $\text{RHS}$, I did the following. $$ \begin{align*} 2^{2n}&=(1+1)^{2n}\\ &=\binom{2n}{0}+\binom{2n}{1}+\binom{2n}{2}+\cdots+\binom{2n}{n}+\cdots+\binom{2n}{2n}. \end{align*} $$ From $\displaystyle \binom{n}{k}=\binom{n}{n-k}$, it follows that $$\begin{align*} 2^{2n}=\left[\binom{2n}{0}+\binom{2n}{2n}\right]+\left[\binom{2n}{1}+\binom{2n}{2n-1}\right]+\cdots+\binom{2n}{n}, \end{align*}$$ but I can't proceed any further.

Answer 1

Let $ n $ be a positive integer greater than $ 1 \cdot $

Separating the odd factors from the even ones in the product $ \prod\limits_{k=1}^{2n}{k} $, gives the following : $$ \prod_{k=1}^{2n}{k}=\prod_{k=1}^{n}{\left(2k\right)}\prod_{k=0}^{n-1}{\left(2k+1\right)} $$

Meaning, we have $ \prod\limits_{k=0}^{n-1}{\left(2k+1\right)}=\frac{\prod\limits_{k=1}^{2n}{k}}{\prod\limits_{k=1}^{n}{\left(2k\right)}}=\frac{\left(2n\right)!}{2^{n}n!}\cdot $

Thus, $$ \frac{\prod\limits_{k=0}^{n-1}{\left(2k+1\right)}}{\prod\limits_{k=1}^{n}{\left(2k\right)}}=\frac{\left(2n\right)!}{2^{2n}\left(n!\right)^{2}}=\frac{1}{2^{2n}}\binom{2n}{n} $$

Answer 2

Here's a direct proof: \begin{align} \binom{2n}{n} &= \frac{(2n)!}{n!^2}\\ &= \frac{\prod_{k=1}^{2n} k}{\prod_{k=1}^n k^2} \\ &= \frac{\left(\prod_{k=1}^n 2k\right)\left(\prod_{k=1}^n (2k-1)\right)}{\prod_{k=1}^n k^2} \\ &= 2^n\prod_{k=1}^n \frac{k (2k-1)}{k^2} \\ &= 2^n\prod_{k=1}^n \frac{2k-1}{k} \\ &= 2^{2n}\prod_{k=1}^n \frac{2k-1}{2k} \end{align}