Question

Prove that $\frac{{}^{4n}{C}_{2n}}{{}^{2n}{C}_n}=\frac{\mathrm{1.3.5......}(4n-1)}{\left\{\mathrm{1.3.5......}\right(2n-1){\}}^2}$

Answer 1

$\frac{{}^{4n}{C}_{2n}}{{}^{2n}{C}_n}=\frac{\left(4n\right)!\left(n\right)!\left(n\right)!}{\left(2n\right)!\left(2n\right)!\left(2n\right)!}$

$=\frac{\left(4n\right)!}{\left(2n\right)!}\times \frac{n!}{\left(2n\right)!}\times \frac{n!}{\left(2n\right)!}$

$=\left[\mathrm{1.3.5......}\right(4n-1\left)\right]\times \frac{1}{\mathrm{1.3.5.....}(2n-1)}\times \frac{1}{\mathrm{1.3.5.....}(2n-1)}$

$=\frac{\mathrm{1.3.5.....}(4n-1)}{\left\{\mathrm{1.3.5....}\right(2n-1){\}}^2}$