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Proof Using General principle of math induction

  1. May 1, 2010 #1
    1. The problem statement, all variables and given/known data

    prove n!>2^n for all n>=4

    2. Relevant equations



    3. The attempt at a solution

    I showed it was true for n=1.

    assume k!>2^k for all k>=4
    then show it for k+1. (k+1)!>=2^(k+1)
    =k!*(k+1)>=2*2^k
    I don't know where to go from here.
     
  2. jcsd
  3. May 1, 2010 #2
    But your base case is n=4!

    Simplify 2*2k
     
  4. May 1, 2010 #3
    sorry, i meant for n=4
     
  5. May 1, 2010 #4

    Mark44

    Staff: Mentor

    This is what you need to show: (k+1)!>=2^(k+1).
    In your next line, though, you are tacitly assuming that this is true. Also, don't connect one inequality to another with = as you are doing here:
    (k+1)!>=2^(k+1)
    =k!*(k+1)>=2*2^k

    You have (k + 1)! = k! * (k + 1) = ? Here's where you use your induction hypothesis (i.e., k! >= 2^k).
     
  6. May 1, 2010 #5
    I guess the factorial is what is throwing me off, I don't know how to use a chain of inequalities that will lead me to something that I can directly compare to 2^k+1 because I don't know how to take the factorial into account or get rid of it.
     
  7. May 1, 2010 #6
    Use (k+1)! = (k+1) k! and apply the induction hypothesis
     
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