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Math Help - Can you help me?

  1. #1
    Junior Member
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    Can you help me?

    Prove that 3^n > n^3 for all positive integers n.

    I first tried n= 1, which showed 3>1
    n= 2, 9>8
    but n = 3 became 27 = 27, which is not true, but everything else works afterwards.

    I was given the hint to use induction. So I tried this:

    3^(n+1) > (n+1)^3
    3*(3^n) > n^3 + 3n^2 + 3n + 1

    What do I do next?
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  2. #2
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    Suppose 3^n > n^3 for some n = k, k \ge 4 which was shown.

    For the inductive step, you want to show that 3^{k+1} > (k+1)^3.

    Well (k+1)^3 = k^3 + 3k^2 + 3k + 1.

    Since it is known that k \ge 4 then k^3 \ge 4k^2 > 3k^2.

    Also, k \ge 4 leads to k^2 \ge 16, so k^2 - 3 \ge 13 and it follows that  k(k^2-3) \ge 52 > 1. Rearranging, it is evident that  3k + 1 < k^3.

    So applying the above revelations, k^3 + 3k^2 + 3k + 1 = (k^3) + (3k^2) + (3k+1)
    < k^3 + k^3 + k^3 = 3k^3.

    It is hypothesized that k^3 < 3^k, so 3k^3 < 3 \cdot 3^k = 3^{k+1}.

    Thus, (k+1)^3 < 3^{k+1}, and by induction, 3^n > n^3 is true for all n \ge 4.
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  3. #3
    Junior Member
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    Jun 2006
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    Thank you for your help. But I need a little more explantion.


    I don't quite know where or why you did this particular line: k^3 >= 4k^2>3k^2

    Then why did you do the two lines after the above line?
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  4. #4
    Member
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    San Diego
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    I'll try to be clearer:

    I wanted to make it clear that k^3 > 3k^2 and k^3 > 3k+1.

    Using both of these facts, then k^3+k^3 > 3k^2 + 3k +1, so k^3+k^3+k^3 > k^3 + 3k^2 + 3k +1 which is the expansion of (k+1)^3.

    So all of the above sets up:

    (k+1)^3 < 3k^3

    and since k^3 < 3^k, then 3k^3 < 3 \cdot 3^k. in other words, (k+1)^3 < 3^{k+1}, which is what we wre trying to show.
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