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Proof Using General principle of math induction

  • Thread starter kolley
  • Start date
  • #1
17
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Homework Statement



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

Homework Equations





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.
 

Answers and Replies

  • #2
614
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I showed it was true for n=1.
But your base case is n=4!

Simplify 2*2k
 
  • #3
17
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sorry, i meant for n=4
 
  • #4
33,174
4,858
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
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).
 
  • #5
17
0
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.
 
  • #6
614
0
Use (k+1)! = (k+1) k! and apply the induction hypothesis
 

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