Q18 of 118 Page 18

Prove that : 4nC2n : 2nCn = [1 . 3 . 5 …. (4n – 1)] : [1.3.5….. (2n – 1)]2.

Given that we need to prove:


4nC2n : 2nCn = [1 . 3 . 5 …. (4n – 1)] : [1.3.5….. (2n – 1)]2.


Consider L.H.S:


We know that


And also n! = n(n – 1)(n – 2)…………2.1


⇒


⇒


⇒


⇒


⇒


⇒


⇒


⇒


⇒


⇒


⇒


= R.H.S


∴ L.H.S = R.H.S, thus proved.


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