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How can I solve this?(2n+1)!/n!x(n+1)! x (2n-1)!/n!(n-1)! < [(2n)!/(n!)^2]^2

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marinaart | Student | eNoter

Posted April 28, 2013 at 3:36 PM via web

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How can I solve this?

(2n+1)!/n!x(n+1)! x (2n-1)!/n!(n-1)! < [(2n)!/(n!)^2]^2

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marinaart | Student | eNoter

Posted April 28, 2013 at 4:10 PM (Answer #1)

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(2n+1)!/n!*(n+1)!*(2n-1)!/n!(n-1)! < [(2n)!/(n!)^2]^2

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pramodpandey | College Teacher | Valedictorian

Posted April 29, 2013 at 6:54 AM (Answer #2)

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`((2n+1)!)/{n!(n+1)!}xx((2n-1)!)/{n!(n-1)!}<(((2n)!)/(n!)^2)^2`

`` LHS.=`{(2n+1)(2n)!}/{n!(n+1)n!}xx{(2n)(2n-1)!}/{(2n)n!(n-1)!}`

`={(2n+1)(2n)!(2n)!}/{(n!)^2 2(n+1)(n!)^2}`

`={(2n+1)((2n)!)^2}/{(2n+2)(n!)^4}`

`={(2n+1)/(2n+2)}(((2n)!)/(n!)^2)^2`

`But`

`2n+1<2n+2`

`(2n+1)/(2n+2)<1`

`((2n+1)/(2n+2))(((2n)!)/(n!)^2)^2< (((2n)!)/(n!)^2)^2`

`LHS<RHS`

proved.

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