Question

Prove that $\frac{\left(2n\right)!}{n}=\{1\times 3\times 5....(2n-1\left)\right\}{2}^n$.

Answer 1

$RHS$

$1\times 3\times \mathrm{5......}(2n-1)$

Multiplying and dividing $2\times 4\times 6\times \mathrm{8........2}n$ and we get,

$\frac{(1\times 3\times \mathrm{5......}(2n-1))\times (2\times 4\times 6\times \mathrm{8.......2}n)}{(2\times 4\times 6\times \mathrm{8.......2}n)}{2}^n$

$\Rightarrow \frac{1\times 2\times 3\times \mathrm{4.......2}n\times (2n-1)}{2^n(1\times 2\times 3\times \mathrm{4.......}n)}{2}^n$

$\Rightarrow \frac{\left(2n\right)!}{2^{n}n!}{2}^n$

$\Rightarrow \frac{\left(2n\right)!}{n!}$

$LHS$

Hence proved.