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I've been stuck on this question for a bit, I am unsure of how to factorize factorial and is stuck at this stage. If someone can give me a hint on where to go next, it would be highly appreciated!

$$\prod_{k=1}^{n} (2k - 1) = \frac{(2n)!}{n! 2^n} \quad (n \geq 1).$$

Where I got to

$\prod_{k=1}^{a+1} (2k-1)$

$= (\prod_{k=1}^{a} (2k-1))(2(a+1)-1)$

$= \frac{(2a)!}{a!2^a}(2a+1)$

$= \frac{(2a)!(2a+1)}{a!2^a}$

Red Five
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Kitty Summoner
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  • Good, then multiply both the denominator and numerator by something that makes the denominator $(a+1)!2^{a+1}$. – peterwhy Apr 28 '24 at 00:48

1 Answers1

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$$= \frac{(2a)!(2a+1)}{a!2^a}$$ $$=\frac{2a+2}{2(a+1)}×\frac{(2a)!(2a+1)}{a!2^a}$$ $$=\frac{(2a+2)!}{(a+1)!2^{a+1}}$$

Gwen
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