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Question

Prove that (2n!)n!=2n(1.3.5....(2n1)).

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Solution

(2n!)=2n(2n1)(2n2)....4.3.2.1
Taking 2 common from alternative even terms,we get
(2n!)=(2.2.....n times) [n(2n1)(n1)....2.3.1.1]
(2n!)=2n[(2n1)(2n3)...3.1][n(n1)...2.1]
(2n!)=2n[(2n1)(2n3)...3.1]n!
(2n!)n!=2n(1.3.5.7...(2n1)]
Hence proved.

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