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I have this formula which seems to work for the product of the first n odd numbers (I have tested it for all numbers from $1$ to $100$):

$\prod_{i = 1}^{n} (2i - 1) = \frac{(2n)!}{2^{n} n!}$

How can I prove that it holds (or find a counter-example)?

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    As a $n$ote: wh$a$t you h$a$ve is sometimes termed $a$s the "double factorial".2011-10-02

3 Answers 3

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The idea is to "complete the factorials":

$ 1\cdot 3 \cdot 5 \cdots (2n-1) = \frac{ 1 \cdot 2 \cdot 3 \cdot 4 \cdots (2n-1)\cdot (2n) }{2\cdot 4 \cdot 6 \cdots (2n)} $

Now take out the factor of $2$ from each term in the denominator:

$ = \frac{ (2n)! }{2^n \left( 1\cdot 2 \cdot 3 \cdots n \right)} = \frac{(2n)!}{2^n n!}$

A mathematician may object that there is a small gray area about what exactly happens between those ellipses, so for a completely rigorous proof one would take my post and incorporate it into a proof by induction.

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For the induction argument, $\begin{align*} \prod_{i=1}^{n+1}(2i-1)&=\left(\prod_{i=1}^n(2i-1)\right)(2n+1)\\ &= \frac{(2n)!(2n+1)}{2^n n!} \end{align*}$ by the induction hypothesis. Now multiply that last fraction by a carefully chosen expression of the form $\dfrac{a}a$ to get the desired result.

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$\Pi_{k=1}^n(2k-1) = \Pi_{k=1}^n(2k-1) \frac{\Pi_{k=1}^n(2k)}{\Pi_{k=1}^n(2k)} =\frac{\Pi_{k=1}^{2n}k}{2^n\Pi_{k=1}^n k} = \frac{(2n)!}{2^n n!}$

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    This is the same answer as the one given **four years ago** by Ragib, only with fewer explanations. It's good that you want to help by answering questions, but maybe you could consider answering some that haven't been answered yet?2015-08-31