I have been set this question and don't know wherto begin to get powers from..
Show that,,$\displaystyle \frac{(2n)!}{n!}=2^n.1.3.5...(2n-1)$
$\displaystyle \left[ {2n} \right]! = \left( {2n} \right)\left( {2n - 1} \right)\left( {2n - 2} \right)\left( {2n - 3} \right)\left( {2n - 4} \right) \cdots (4)(3)(2)(1)$
$\displaystyle \left[ {2n} \right]! = 2\left( n \right)\left( {2n - 1} \right)2\left( {n - 1} \right)\left( {2n - 3} \right)2\left( {n - 2} \right) \cdots 2(2)(3)2(1)(1)$, factor out 2's.
$\displaystyle \left[ {2n} \right]! = 2^n \left( n \right)\left( {2n - 1} \right)\left( {n - 1} \right)\left( {2n - 3} \right)\left( {n - 2} \right) \cdots (2)(3)(1)(1)$, there are n 2's. Do you see a factor of n!?
$\displaystyle \left[ {2n} \right]! = 2^n \left( {2n - 1} \right)\left( {2n - 3} \right)\left( {n - 2} \right) \cdots (3)(1)(n!)$