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Homework Help: 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


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    be the statement

    3^n > n^3

    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].


    3^{3+1} = 3 \cdot 3 ^k > 3 \cdot k^3 = k^3 + k^3 + k^3

    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
    k^3 = k \cdot k^2 > 3k^2
    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:

    For n+1=

    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

    from 1 and 2


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


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    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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