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Math Help - Proof by induction, involving factorials

  1. #1
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    Proof by induction, involving factorials

    Prove that 2! \cdot 4! \cdot 6!.....(2n)! \geq \left\[(n+1)!]^n
    I've already proved that P(1) is a true statement.

    My inductive hypothesis is that 2! \cdot 4! \cdot 6!.....(2k)! \geq \left\[(k+1)!]^k is true.

    I'm trying to show that P(k+1) is true: 2! \cdot 4! \cdot 6!.....(2k+2)! \geq \left\[(k+2)!]^{k+1}

    2! \cdot 4! \cdot 6!.....(2k+2)!
    \Rightarrow 2! \cdot 4! \cdot 6!.....(2k+2)(2k+1)(2k)!
    \Rightarrow 2! \cdot 4! \cdot 6!.....(2k)!(2k+2)(2k+1)

    I'm stuck here though. How can this lead to being \geq \left\[(k+2)!]^{k+1}?
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  2. #2
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    Re: Proof by induction, involving factorials

    2!\cdot 4! \cdot 6! \cdots (2k)!\cdot (2k+2)! \ge [(k+1)!]^k(2k+2)! by the induction hypothesis.

    \begin{align*}[(k+1)!]^k(2k+2)! & = [(k+1)!]^k(k+1)!(k+2)\cdots (2k+2) \\ & = [(k+1)!]^{k+1}\underbrace{(k+2)\cdots (2k+2)}_{k+1\text{ terms}}\\ & \ge [(k+1)!]^{k+1}\underbrace{(k+2)\cdots (k+2)}_{k+1\text{ times}} \\ & = [(k+2)!]^{k+1}\end{align*}
    Last edited by SlipEternal; November 12th 2013 at 07:53 AM.
    Thanks from topsquark and MadSoulz
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  3. #3
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    Re: Proof by induction, involving factorials

    Quote Originally Posted by SlipEternal View Post
    2!\cdot 4! \cdot 6! \cdots (2k)!\cdot (2k+2)! \ge [(k+1)!]^k(2k+2)! by the induction hypothesis.

    \begin{align*}[(k+1)!]^k(2k+2)! & = [(k+1)!]^k(k+1)!(k+2)\cdots (2k+2) \\ & = [(k+1)!]^{k+1}\underbrace{(k+2)\cdots (2k+2)}_{k+1\text{ terms}}\\ & \ge [(k+1)!]^{k+1}\underbrace{(k+2)\cdots (k+2)}_{k+1\text{ times}} \\ & = [(k+2)!]^{k+1}\end{align*}
    I'm assuming my  (2k+1) isn't correct because that would give an odd number.

    Are you pulling out (k+1)! from (2k+2)!? In the second line.
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