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Math Help - Prove by Induction

  1. #1
    Member kjchauhan's Avatar
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    Prove by Induction

    Please help me to solve the following

    Question : Prove by Induction,

    \frac{(2n)!}{2^{2n}(n!)^{2}} \le \frac{1}{\sqrt{3n+1}} , n\in N.

    I tried and prove for n=1 and 2 and assumed for n=k, but I couldn't get for n=k+1, I got the result as,

    \frac{(2(k+1))!}{2^{2(k+1)}((k+1)!)^{2}} = \frac{(2k)!}{2^{2k}(k!)^{2}} \into \frac{(2k+1)}{2(k+1)}

    now How can we say that the right hand sid of the above result is \le \frac{1}{\sqrt{3(k+1)+1}}


    so help me to solve this
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  2. #2
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    Re: Prove by Induction

    Quote Originally Posted by kjchauhan View Post
    Question : Prove by Induction,
    \frac{(2n)!}{2^{2n}(n!)^{2}} \le \frac{1}{\sqrt{3n+1}} , n\in N.
    I have no idea if this can help you, but here it is.

    \frac{(2n)!}{2^n(n!)}=(2n-1)(2n-3)\cdots(3)(1)
    Thanks from wilhelm
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