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Math Help - mathematical induction again

  1. #1
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    mathematical induction again

    I need to use mathematical induction to prove 2^n<n! for all n>=4. The base case is easy, and in the inductive step I get to 2(2^n)<(n+1)n! But am I going in the wrong direction, or am I just missing something?
    Last edited by tottenc; July 30th 2007 at 05:12 PM.
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  2. #2
    Senior Member tukeywilliams's Avatar
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    Assume  2^k < k! for  k \geq 4 . Then  2^{k+1} < 2k! < (k+1)! = k!(k+1) .  k+1 > 2 for  k >1 . Hence  2^{k+1} < (k+1)! for  k \geq 4 and we have proved the inductive step.
    Last edited by tukeywilliams; July 30th 2007 at 11:22 AM.
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  3. #3
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    Sorry, stupid typo
    2^n<n!<br />
for all n>=4
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  4. #4
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    Assume that N \ge 4\quad \& \quad 2^N  < N! for the inductive step.

    Then 2^{N + 1}  = 2\left( {2^N } \right) < \left( {N + 1} \right)\left( {N!} \right) = \left( {N + 1} \right)!
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  5. #5
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    The answer to these things is always staring me right in the face, and I just can't see it until someone points it out. Thanks again
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  6. #6
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by tottenc View Post
    The answer to these things is always staring me right in the face, and I just can't see it until someone points it out. Thanks again
    It just takes some practice.

    -Dan
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