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Stuck on Proof by induction of 2^n>n^3 for all n>=10

  1. Nov 5, 2015 #1
    1. The problem statement, all variables and given/known data
    Using the principle of mathematical induction, prove that for all n>=10, 2^n>n^3

    2. Relevant equations

    2^(n+1) = 2(2^n)
    (n+1)^3 = n^3 + 3n^2 + 3n +1

    3. The attempt at a solution
    i) (Base case) Statement is true for n=10
    ii)(inductive step) Suppose 2^n > n^3 for some integer >= 10
    (show that 2^(n+1) > (n+1)^3 )
    Consider 2^(n+1).
    2^(n+1)= 2(2^n) > 2(n^3) = n^3 + n^3
    (Ok, so this is where i'm stuck. Can I say n^3 > 3n^2 + 3n +1 because n>=10? Because if I can say that, then I can proceed with n^3 + n^3 > n^3 + 3n^2 + 3n +1 = (n+1)^3. I just don't know if i have to further justify it. Should I do another proof by induction to show that n^3 > 3n^2 + 3n +1 for n>=10? Or can I make a general statement that the power of 3 is higher than a power of 2 and so on)

    Thank you!
  2. jcsd
  3. Nov 5, 2015 #2


    Staff: Mentor

    3n^2 + 3n + 1 < 3n^2 + 3n^2 + 3n^2 = 9n^2 < 10n^2 <= n^3 should do
  4. Nov 5, 2015 #3


    User Avatar
    Staff Emeritus
    Science Advisor

    n^3- 3n^2+ 3n- 1 has derivative 3n^2- 6n+ 3= 3(n^2- 2n+ 1)= 3(n- 1)^2. That has a single 0 at n= 1< 10. Further at n= 10, 3(n-1)^2= 3*81> 0 so n^3- 3n^2+ 3n- 1 is increasing for all n> 10 and when n= 10, its value is 1000- 300+ 30+ 1= 729> 0 so n^3- 3n^2+ 3n- 1> 0 for all n> 10. Yes, n^3> 3n^2- 3n+ 1 for n> 10.
  5. Nov 5, 2015 #4
    Wow. That's creative. So basically I can't just say that n^3 > 3n^2 + 3n + 1 because 3n^2 isn't the only term? And is there a reason your last inequality is 10n^2 <= n^3 instead of just 10n^2 < n^3? Can I just say 10n^2 <n^3 because n>=10?

    Thank you!
  6. Nov 5, 2015 #5


    Staff: Mentor

    I first had 10n^2 < n^3 but this is not true for the possible n=10. So either you state n>10 (which can be done in the induction step) or you just write the line from right to left. One "<" is enough.
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