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Prove by induction

  1. Sep 15, 2009 #1
    3n>n3 where n >3

    I know I have to use proof by induction to solve this.
    assume for f(4)is true
    for f(n+1)=3n+1>3n3

    However after that I dont have a clue of how to getting it into (n+1)3
    Any hints will be greatly apperciated
     
  2. jcsd
  3. Sep 15, 2009 #2

    statdad

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    Homework Helper

    Let

    [tex]
    A_n
    [/tex]

    be the statement

    [tex]
    3^n > n^3
    [/tex]

    You want to show [tex] A_n [/tex] is true for [tex] n \ge 4 [/tex].

    It shouldn't be hard to show [tex] A_4 [/tex] is true. Assume it is true for [tex] k \ge 4 [/tex].

    Now

    [tex]
    3^{3+1} = 3 \cdot 3 ^k > 3 \cdot k^3 = k^3 + k^3 + k^3
    [/tex]

    Now play with the terms on the right, making the new expressions ever smaller, to build up to the expansion of [tex] (k+1)^3 [/tex].

    (Hint for a start: You know by hypothesis [tex] k \ge 4 > 3 [/tex], so
    [tex]
    k^3 = k \cdot k^2 > 3k^2
    [/tex]
    which is the second term in the expansion of [tex] (k+1)^3 [/tex])
     
    Last edited by a moderator: Sep 15, 2009
  4. Sep 16, 2009 #3
    Sorry for the late reply
    This is th best I could get at:

    3n>n3
    For n+1=
    3n+1>3n

    given n>3 , I found the following:

    • n3>3n2
    • n2>3n
    • n-2>1
    with the best intentions I did the following:

    Added all them up and got
    n3>3n2+3n+1-n2-n+2--(1)
    3n+1-2n3>n3--(2)

    from 1 and 2
    3n+1-2n3>3n2+3n+1-n2-n+2

    3n+1+n2+n-2-n3>(n+1)3

    But after that im stuck it would be great some one could help. thnx
     
  5. Sep 16, 2009 #4
    anyone?
     
  6. Sep 16, 2009 #5

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Statdad showed that you could get to [itex]k^3+ k^3+ k^3[/itex] and you want to compare that to [itex](k+1)^3= k^3+ 3k^2+ 3k+ 1[/itex]. Can you show that [itex]k^3> 3k^2[/itex] when k> 3? Can you show that [itex]k^3> 3k+ 1[/itex] when k> 3?
     
  7. Sep 17, 2009 #6
    nope I cant .:cry:
     
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