4nC2n : 2nCn = [1.3.5...(4n-1)] : [1.3.5...(2n-1)]^2
Answers
= (2n)! / (n!)2
= (2n) ( 2n - 1) (2n - 2) -----------4.3.2.1 / (n!)2
= 2xn ( 2n - 1) 2 ( n - 1) -----------------2x2.3.2x1.1 / (n!)2
= 2n [n(n - 1)(n - 2)(n -3) ----------3 x2 x1][(2n -1)(2n -3) ---------------5x3x1] / (n!)2
= 2n n! [(2n -1)(2n -3) ---------------5x3x1] / (n!)2
= 2n [(2n -1)(2n -3) ---------------5x3x1] / (n!)
= 2n[ 1x3x5 -----------------(2n -3)(2n -1)] / (n!)
Answer:
4nC2n:2nCn=[1.3.5.....(4n-1)]:[1.3.5.....(2n-1)]
from LHS
4nC2n :2ncn=4nC2n/2nCn
=4n!/(4n-2n)! 2n! /2n! /(2n-n)!n!
=4n! n! n! /2n! 2n! 2n!
=[4n(4n-1)(4n-2)......3.2.1] n! n!
2n! 2n! 2n!
=[1.3.5...(4n-1) ](2.4.6.....4n)n!n!
2n! 2n! 2n!
=[1.3.5...(4n-1)] 2^(2n)*(1.2.3......2n)n! n!
2n! 2n! 2n!
=2^(2n)[1.3.5...(4n-1)] 2n! n! n!
2n! 2n! 2n!
=2^(2n)[1.3.5...(4n-1)]n! n!
2n! 2n!
= 2^(2n)[1.3.5...(4n-1)]n! n!
[1.3.5...(2n-1)]^2*(2.4.6.....2n)^2
= 2^(2n)[1.3.5...(4n-1)]n! n!
[1.3.5...(2n-1)]^2*2^2n[1.2.3....n]^2
= 2^(2n)[1.3.5...(4n-1)]n! n!
[1.3.5...(2n-1)]^2*2^(2n)*n!n!
= [1.3.5...(4n-1)] =RHS
[1.3.5...(2n-1)]^2
hence proved.
