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Proof of a Sequence

  1. Nov 1, 2004 #1

    JasonRox

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    I am writing the exact problem.

    1. Prove that for all natural numbers n the following equality is satisfied:

    [tex]1^3 + 2^3 + 3^3 + ... + n^3 = \frac{n^2(n+1)^2}{4}[/tex]

    I am doing the exact method they used in the book, which I don't entirely agree on.

    My answer is different than the one in the back of the book, but they come to the same conclusion.

    Let n = k + 1,

    [tex]\frac{(k+1)^2(k+1+1)^2}{4} = \frac{k^4+6k^3+13k^3+12k+4}{4} = \frac{k^2(k+1)^2}{4} + (k+1)^3[/tex]

    Since, [itex]\frac{k^2(k+1)^2}{4}[/itex] is true by our assumption then adding [itex](k+1)^3[/itex] does not change the solution when n = k + 1.

    I left some of the calculations out so I don't have to waste an hour doing tex, but I don't think there is any mistakes in it.

    Obviously, if you test it, it is right.

    Does that constitute as a proof?

    I don't agree with the way they do it because they are assuming that there assumption is true. I really don't know.

    Any comments?
     
  2. jcsd
  3. Nov 1, 2004 #2

    arildno

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    So, what you're objecting to is their expansion of what ought to be the (k+1)-sum?
     
  4. Nov 1, 2004 #3

    arildno

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    I can't really agrree to the objection:
    They are using the "k-true" assumption on their right-hand side of an equality.
    By noting therefore that the right-hand side is nothing but the k+1-sum, they've proven the induction step in an admittedly roundabout, but, IMO, valid way.

    There is NO assumption lying in expanding the ("arbitrarily chosen" ) expression on the left-hand side!
     
  5. Nov 2, 2004 #4

    Gokul43201

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    JasonRox, do you know how inductive proofs work ?
     
  6. Nov 2, 2004 #5
    The problem here, I feel, is that Jason Rox left out a very, very important fact: he failed to consider proving the formula for a small value, here k=1. I think that is why he does not see the induction. For k=1, the formula gives:

    [tex]1^3=1=\frac{1^2*2^2}{4}[/tex]

    If it is true for k it is also true for k+1, and it is shown true for 1; thus since it is true for k=1, so it is also true for k+1 = 2, and if it is true for k=2 then it is true for k=3, and so on and so on.
     
    Last edited: Nov 2, 2004
  7. Nov 2, 2004 #6

    JasonRox

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    Sorry, I forgot to mention that it works for k=1.

    I understand why that is necessary. If it didn't work for one, then it wouldn't work for all natural numbers u_n if u_1 + u_2 + ... + u_n.

    Note for Gokul: I have no clue how inductive proofs work. I'll make a note to read about it in one of my books.
     
  8. Nov 2, 2004 #7
    ... :wink:
     
  9. Nov 2, 2004 #8

    JasonRox

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    I thought about it in that fashion, but I just didn't think it was enough.

    It makes sense if you think about it. I haven't done proofs in this fashion before.
     
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