Question

Express 1.3.5......(2n-1) as a fraction whose numerator is 2n!

Answer 1

Sol:

Given 2nCn =  (2n)! / n! n!

= (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!)