Prove :$\displaystyle \mathop \prod \limits_{i = 2}^n \left( {2i - 1} \right) = \frac{{\left( {2n} \right)!}}{{2^n \times n!}}$$\displaystyle 2. \displaystyle \prod_{i=2}^n(2i-1)=3\cdot5\cdot7\cdot\ldots\cdot(2n-1)= \displaystyle =\frac{1\cdot 2\cdot 3\cdot 4\cdot 5\cdot\ldots\cdot(2n)}{2\cdot 4\cdot 6\ldots\cdot(2n)}=\frac{(2n)!}{2^n(1\cdot 2\cdot 3\ldots\cdot n)}=\frac{(2n)!}{2^n\cdot n!} 3. Originally Posted by dhiab Prove :\displaystyle \mathop \prod \limits_{i = 2}^n \left( {2i - 1} \right) = \frac{{\left( {2n} \right)!}}{{2^n \times n!}}$$\displaystyle$
Assume $\displaystyle \mathop \prod \limits_{i = 2}^n \left( {2i - 1} \right) = \frac{{\left( {2n} \right)!}}{{2^n \times n!}}$
we must show $\displaystyle \mathop \prod \limits_{i = 2}^{n+1} \left( {2i - 1} \right) = \frac{{\left( {2(n+1)} \right)!}}{{2^{n+1} \times (n+1)!}}$
$\displaystyle \mathop \prod \limits_{i = 2}^{n+1} \left( {2i - 1} \right) = $$\displaystyle \mathop (\prod \limits_{i = 2}^{n} \left( {2i - 1} \right) )(2(n+1)-1)= \displaystyle \frac{2n!}{2^n\cdot n!}(2n+1)=$$\displaystyle \frac{2n!}{2^n\cdot n!}(2n+1)\left(\frac{2(n+1)}{2(n+1)}\right)=$ $\displaystyle =$$\displaystyle \frac{{\left( {2(n+1)} \right)!}}{{2^{n+1} \times (n+1)!}}$