1. ## Factoriel : 1

Prove : $\mathop \prod \limits_{i = 2}^n \left( {2i - 1} \right) = \frac{{\left( {2n} \right)!}}{{2^n \times n!}}$ $
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2. $\prod_{i=2}^n(2i-1)=3\cdot5\cdot7\cdot\ldots\cdot(2n-1)=$

$=\frac{1\cdot 2\cdot 3\cdot 4\cdot 5\cdot\ldots\cdot(2n)}{2\cdot 4\cdot 6\ldots\cdot(2n)}=\frac{(2n)!}{2^n(1\cdot 2\cdot 3\ldots\cdot n)}=\frac{(2n)!}{2^n\cdot n!}$

3. Originally Posted by dhiab
Prove : $\mathop \prod \limits_{i = 2}^n \left( {2i - 1} \right) = \frac{{\left( {2n} \right)!}}{{2^n \times n!}}$ $
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Induction, I trust you can do the base case.

Assume $\mathop \prod \limits_{i = 2}^n \left( {2i - 1} \right) = \frac{{\left( {2n} \right)!}}{{2^n \times n!}}$

we must show $\mathop \prod \limits_{i = 2}^{n+1} \left( {2i - 1} \right) = \frac{{\left( {2(n+1)} \right)!}}{{2^{n+1} \times (n+1)!}}$

$\mathop \prod \limits_{i = 2}^{n+1} \left( {2i - 1} \right) =$ $\mathop (\prod \limits_{i = 2}^{n} \left( {2i - 1} \right) )(2(n+1)-1)=$ $\frac{2n!}{2^n\cdot n!}(2n+1)=$ $\frac{2n!}{2^n\cdot n!}(2n+1)\left(\frac{2(n+1)}{2(n+1)}\right)=$ $=$ $\frac{{\left( {2(n+1)} \right)!}}{{2^{n+1} \times (n+1)!}}$