Question

Prove that: ⁴ⁿC_2n : ²ⁿC_n = [1 · 3 · 5 ... (4n − 1)] : [1 · 3 · 5 ... (2n − 1)]².

Answer 1

$$\begin{gathered} \frac{{}^{4n}{C}_{2n}}{{}^{2n}{C}_n}=\frac{1.3.5...\left(4n-1\right)}{{\left[1.3.5...\left(2n-1\right)\right]}^2} \\ \mathrm{LHS}=\frac{{}^{4n}{C}_{2n}}{{}^{2n}{C}_n} \\ =\frac{\left(4n\right)!}{\left(2n\right)!\left(2n\right)!}\times \frac{n!n!}{\left(2n\right)!} \\ =\frac{\left[4n\times \left(4n-1\right)\times \left(4n-2\right)\times \left(4n-3\right).................3\times 2\times 1\right]\times {\left(n!\right)}^2}{{\left[2n\times \left(2n-1\right)\times \left(2n-2\right).......3\times 2\times \times 1\right]}^2\left(2n\right)!} \\ =\frac{\left[1\times 3\times 5........\left(4n-1\right)\right]\left[2\times 4\times 6...............4n\right]\times {\left(n!\right)}^2}{{\left[1\times 3\times 5\times .........\left(2n-1\right)\right]}^2{\left[2\times 4\times 6\times .......2n\right]}^2\times \left(2n\right)!} \\ =\frac{\left[1\times 3\times 5........\left(4n-1\right)\right]\times 2^{2n}\times \left[1\times 2\times 3..........2n\right]{\left(n!\right)}^2}{{\left[1\times 3\times 5\times .........\left(2n-1\right)\right]}^2\times 2^{2n}{\left[1\times 2\times 3\times .......n\right]}^2\left(2n\right)!} \\ =\frac{\left[1\times 3\times 5........\left(4n-1\right)\right]\left(2n\right)!{\left[n!\right]}^2}{{\left[1\times 3\times 5\times .........\left(2n-1\right)\right]}^2{\left[n!\right]}^2\left(2n\right)!} \\ =\frac{\left[1\times 3\times 5........\left(4n-1\right)\right]}{{\left[1\times 3\times 5\times .........\left(2n-1\right)\right]}^2}=\mathrm{RHS} \\ \mathrm{Hence},\mathrm{proved}. \end{gathered}$$