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Induction proof of 2^n < n!

  1. Jan 11, 2005 #1
    I posted this elsewhere but realized it ought to be in the homework section. I have to use induction to prove that for n>=4, 2^n < n! is true, but I don't know wehre to start. I have the base case proven, but then I don't know where to go after I have my Inductive Hypothesis that it works for all n's greater than 4. Any help would be very appreciated. Thank you
    Josh
     
  2. jcsd
  3. Jan 11, 2005 #2
    NateTG helped me in the original posting, so you can disregard this (unless you want to do it for fun!) Thanks
    Josh
     
  4. Jan 11, 2005 #3

    Andrew Mason

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    Write out the expression for n!

    [tex]n! = n(n-1)(n-2)(n-3)...(n-(n-1))[/tex]

    Then subtract each term by a positive number so that each term in the product is equal to 2.

    AM
     
  5. Jan 11, 2005 #4

    Curious3141

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    Induction is the easiest way, I think.

    EDIT : Sorry, I just saw the title of your post, and you wanted the induction proof. :smile:

    The initial verification for the case of 4 is easy. Say the inequality holds for some [itex]k[/itex].

    Then [tex]2^{k+1} = 2.2^k < 2.k! < (k+1).k! = (k+1)![/tex] because [itex](k+1) > 2 [/itex] for all [itex]k > 1[/itex].

    And you're done.
     
    Last edited: Jan 11, 2005
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