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Prove that .^(2n)C(n)=(2^(n)xx[1*3*5...(...

Prove that `.^(2n)C_(n)=(2^(n)xx[1*3*5...(2n-1)])/(n !)`.

Text Solution

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`.^(1n)C_(n)=((2n)!)/((n!)(2n-n)!)=((2n)!)/((n!)^(2))`
`=((2n)(2n-1)(2n-2)...4*3*2*1)/((n!)^(2))`
`=([(2n)(2n-2)(2n-4)...4*2]xx[(2n-1)(2n-3)...5*3*1])/((n!)^(2))`
`=(2^(n)[n(n-1)(n-2)...2*1]xx[(2n-1)(2n-3)...5*3*1])/((n!)^(2))`
`=(2^(n)xx(n!)xx[1*3*5...(2n-3)(2n-1)])/((n!)^(2))`
`=(2^(n)xx[1*3*5...(2n-3)(2n-1)])/((n!))`.
Hence, `.^(2n)C_(n)=(2^(n)xx{1xx3xx5xx...xx(2n-1)})/(n!)`.
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