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Proving n^n > 2^n *n! using the Binomial theorem

  1. Sep 2, 2012 #1
    1. The problem statement, all variables and given/known data
    Prove that [itex] n^n > 2^n * n! [/itex] when n > 6 using the Binomial theorem.
    I just proved the Binomial theorem using induction which was not that difficult but in applying what I learned through it's proof I am having difficulty.

    2. Relevant equations
    Binomial theorem = [itex] (x+y)^n = \sum_{k=0}^n\binom{n}{k}x^{n-k}y^k [/itex]


    3. The attempt at a solution
    I attempted setting n= (x+y) to convert the left side of the equation into the form of the binomial theorem, as well as turning the right hand side into the form of the binomial theorem by setting x+y = 2 both to no avail. Actually the "closest" (I put this in quotes because as I couldn't solve it, I have no idea how close I really was) I got was by using induction and turning the equation into [itex]{\frac{(n+1)^n }{2}}= 2^n + n![/itex]
    Thanks for the help guys.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 2, 2012 #2
    Actually, the last equation I wrote have n!/2
    thanks
     
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