Prove that `((2n)!)/(n !)=2^n{1. 3. 5 (2n-1)}`

Explanation

RHS =`(1.3.5...(2n−1))2^n`

Multiply and divide by 2.4.6...2n on RHS


`[1.3.5...(2n−1)]2^n=(1.3.5...(2n−1))2^xx(2.4.6....2n)/(2.4.6...2n)`
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=`([1.3.5...(2n−1)]2^n)/(2.4.6....2n)`
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=`([(2n)(2n−1)....5.4.3.2.1]2^n)/(2.4.6....2n)`

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=`((2n)!2^n)/(2.4.6....2n)`

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