Proving $\prod_{i=1}^n (2i-1)$ = $\frac{(2n)!}{2^nn!}$ for all natural numbers

sequences-and-series products

I've tried to simplify the right side and eliminate the $n!$ in the denominator to see if that helped to make both sides the same, but that didn't work. I've also tried to expand the left side and look for a pattern I could use.

Any ideas? Any and all help is appreciated.


Solution 1:

We have,

$\displaystyle\prod^{\displaystyle\,n}_{\displaystyle\,i=1}(2i-1)$

$=1\cdot3\cdot5\cdot7\cdots(2n-1)$

$=\dfrac{1\cdot3\cdot5\cdot7\cdots(2n-1)\cdot2\cdot4\cdot6\cdots2n}{2\cdot4\cdot6\cdots2n}$

$=\dfrac{1\cdot2\cdot3\cdot4\cdots\,2n}{2^n(1\cdot2\cdot3\cdots\,n)}$

$=\dfrac{(2n)!}{2^n\cdot\,n!}$

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