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I need help with this induction proof

  1. Dec 11, 2012 #1
    1. The problem statement, all variables and given/known data

    Prove that 2^n > n^3 for every integer n >= 10

    2. Relevant equations

    a) Prove for 10,
    b) assume for k
    c) inductive step for k+1

    3. The attempt at a solution

    a)
    2^10 = 1024
    10^3 = 1000
    1024>1000

    b)Assume
    2^k > k^3

    c) 2^(k+1) > (k+1)^3
    (2)(2^k) > (k+1)^3
    (2^k) > ((k+1)^3)/2

    and from here I'm having trouble solving for k >= 10.
    Any help would be appreciated, thanks.
     
  2. jcsd
  3. Dec 12, 2012 #2

    Mark44

    Staff: Mentor

    The inequality above is what you need to show. You can't just assert it.
    Your work should look like this:
    2k+1 = 2 * 2k > 2 * k3 ...

    You then need to show that 2k3 > (k + 1)3, at least for k ≥ 10.

    It suffices to show that 2k3/(k + 1)3 > 1, which can be done using calculus.
     
    Last edited: Dec 12, 2012
  4. Dec 12, 2012 #3
    Can you elaborate on how its enough to show (2k^3)/(k+1)^3 >1 ? I thought you needed to end up at k >=10?
     
  5. Dec 12, 2012 #4

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    You showed that the inequality holds for n = 10.

    Then you show that if the inequality holds for n = k, where k is any number greater than or equal to 10, then the inequality holds for n = k+1 .

    By doing this, you show that if it's true for n=10, then it's true for n=11 .
    If it's true for n=11, then it's true for n=12 .

    If it's true for n=12, then it's true for n=13 .

    If it's true for n=13, then it's true for n=14 .

    If it's true for n=14, then it's true for n=15 .

    ...​
    So assume that 2k > k3, for some integer k ≥ 10 .

    From that show that 2k+1 > (k+1)3 .
     
  6. Dec 12, 2012 #5

    Mark44

    Staff: Mentor

    (2k3)/(k+1)3 >1 is equivalent to 2k3 > (k + 1)3.
     
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