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Mathematical Induction trouble with 1 step!

  1. Jan 21, 2006 #1
    Hi there folks, I have just a small problem with a specific induction problem. The problem itself is: "Prove [tex]n! > 4^n[/tex], for all n >= 9."

    So here's my work:

    1) Show true for n = 9
    LS
    9! = 362880

    RS
    4^9 = 262144

    .:. LS > RS

    2) Assume true for n = k
    i.e. Assume that k! > 4^k

    3) Prove true for n = k+1
    i.e. Prove that (k+1)! > 4^(k+1)

    So I begin to expand the LHS out

    (k+1)! = (k+1)(k)!
    > (K+1)(4^k) (by induction hypothesis)


    this is the problem that I encounter. I get stuck here because I don't exactly know how to followthrough at this point. How does (k+1)(4^k) become greater that 4^(k+1) ?
     
  2. jcsd
  3. Jan 21, 2006 #2

    VietDao29

    User Avatar
    Homework Helper

    If k >= 9, then what can you say about k + 1? How is k + 1 compared to 4 (greater, less than, or equal to)?
    So from there can you prove that (k + 1)4k > 4k + 1 = 4 4k?
     
  4. Jan 22, 2006 #3
    OH. I see, since k always going to be at least 10. I didn't see it that way.

    Dammit! Oh well, a few silly mistakes cost me quite alot.

    Thanks.
     
    Last edited: Jan 22, 2006
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