$\frac{4n_{C_{2n}}}{2n_{C_n}}=\_\_\_\_\_\_\_\_\_\_\_\_\_$

$$\begin{gathered} \frac{4_{n_{C_{2n}}}}{2n_{C_n}}=\frac{\left(4n\right)!}{(2n!)}\times \frac{{(n!)}^2}{2n!} \\ =\frac{[\mathrm{1.2.3.4.5.6....}(4n-2).(4n-1)\left(4n\right)]{(n!)}^2}{\left[\mathrm{1.2.3.4.5......}(2n-2)(2n-1).2n\right]2n!} \\ =\frac{\left[\mathrm{2.4.6.8........}(4n-2).4n\right]\left[\mathrm{1.3.5......}(4n-1)\right](n!)}{{\left[\mathrm{1.3.5....}(2n-1)\right]}^2{\left[\mathrm{2.4.6.8.....2}n\right]}^2(2n!)} \\ =\frac{\cancel{2^{2n}}.\cancel{2n!}\left[\mathrm{1.3.5.....}(4n-1)\right]\cancel{{(n!)}^2}}{{\left[\mathrm{1.3.5....}(2n-1)\right]}^2.\cancel{2^{2n}}\cancel{{(n!)}^2}\cancel{2n!}} \\ =\frac{\mathrm{1.3.5....}(4n-1)}{{\left[\mathrm{1.3.5.....}(2n-1)\right]}^2} \end{gathered}$$