Prove that:
Question:

Prove that: 4nC2n : 2nCn = [1 · 3 · 5 … (4n − 1)] : [1 · 3 · 5 … (2n − 1)]2.

Solution:

$\frac{{ }^{4 n} C_{2 n}}{{ }^{2 n} C_{n}}=\frac{1.3 .5 \ldots(4 n-1)}{[1.3 .5 \ldots(2 n-1)]^{2}}$

$\mathrm{LHS}=\frac{{ }^{4 n} C_{2 n}}{{ }^{2 n} C_{n}}$

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

$=\frac{[4 n \times(4 n-1) \times(4 n-2) \times(4 n-3) \ldots \ldots \ldots \ldots \ldots . \ldots 3 \times 2 \times 1] \times(n !)^{2}}{[2 n \times(2 n-1) \times(2 n-2) \ldots \ldots . .3 \times 2 \times \times 1]^{2}(2 n) !}$

$=\frac{[1 \times 3 \times 5 \ldots \ldots \ldots(4 n-1)][2 \times 4 \times 6 \ldots \ldots \ldots \ldots .4 n] \times(n !)^{2}}{[1 \times 3 \times 5 \times \ldots \ldots \ldots .(2 n-1)]^{2}[2 \times 4 \times 6 \times \ldots \ldots .2 n]^{2} \times(2 n) !}$

$=\frac{[1 \times 3 \times 5 \ldots \ldots \ldots(4 n-1)] \times 2^{2 n} \times[1 \times 2 \times 3 \ldots \ldots \ldots . .2 n](n !)^{2}}{[1 \times 3 \times 5 \times \ldots \ldots \ldots .(2 n-1)]^{2} \times 2^{2 n}[1 \times 2 \times 3 \times \ldots \ldots . n]^{2}(2 n) !}$

$=\frac{[1 \times 3 \times 5 \ldots \ldots . .(4 n-1)](2 n) ![n !]^{2}}{[1 \times 3 \times 5 \times \ldots \ldots \ldots(2 n-1)]^{2}[n !]^{2}(2 n) !}$

$=\frac{[1 \times 3 \times 5 \ldots \ldots .(4 n-1)]}{[1 \times 3 \times 5 \times \ldots \ldots \ldots(2 n-1)]^{2}}=\mathrm{RHS}$

Hence proved.

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