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Question

Prove that : 4nC2n:2nCn=[1.3.5...(4n1)]:[1.3.5....(2n1)]2.


Solution

4n!(2n)!(2n)!2n!n!n!(nCr=n!r!(nr)!)=(4n!)×(n!)2(2n!)(2n!)×(2n!)2=[1.2.3.4...(4n1)(4n)](n!2)(2n)![1.2.3.4....(2n2)(2n1)(2n)]2=[1.3.5....(4n1)]×[2.4.6...4n]×(n!)2(2n)![1.3.5...(2n1)]2×[2.4.6....(2n2)(2n)]2=[1.3.5...(4n1)]×22n×[1.2.3...2n]×n!2(2n)!×[1.3.5...(2n1)2×22n×n!2]=[1.3.5...(4n1)][1.3.5...(2n1)]2
                                                                  Hence proved.

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