Question:
prove that :- 4n^C 2n : 2n^C n = [1.3.5...(4n-1)] : [1.3.5...(2n-1)]^2
Answer:
Given 4nC2n : 2nCn
= 4nC2n / 2nCn
= [(4n)!/{(4n - 2n)! *(2n)!}]/[(2n)!/{(2n - n)! *(n)!}]
= [(4n)!/{(2n)! *(2n)!}]/[(2n)!/{n! *n!}]
= {(4n)! * (n!)2 }/{(2n)!}3
= [22n * (2n)! *{1*3*5*.............(4n - 1)}*(n!)2 ]/[(2n)! *(2n)!}2 ]
= [(2n )2 * (2n)! *{1*3*5*.............(4n - 1)}]/[(2n)! * (2n )2 *{1*3*5*........(2n - 1)}2 ]
= {1*3*5*.............(4n - 1)}/{1*3*5*........(2n - 1)}2
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