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First attempt at a proof

  1. Dec 15, 2005 #1
    hey everyone,
    I recently bought the the Dover Series book on Number Theory, and the 2nd example on page 5 asks your to prove
    [tex]1^3 + 2^3 + 3^3 ... + n^3 = (1 + 2 + 3....)^2 [/tex]

    Now, we've already proved that [tex]S_n = \frac{n(n+1)}{2}[/tex]

    So here's how I proved it...
    [tex]
    (S_n)^2 = (\frac{n(n+1)}{2})^2 [/tex]

    before we proved how [tex] S_k+1 = S_k + (k + 1) [/tex]

    Which lead me to...
    [tex] \frac{1}(k+1)^2((k+1)+1)^2{4} + (k+1)^2 [/tex]
    [tex] = \sqrt{\frac{(k+1)^2((k+1)+1)^2}{4} + (k+1)^2} [/tex]
    [tex] = \frac{(k+1)((k+1)+1)}{2} + (k+1) [/tex]
    therefor...
    [tex] S_k+1 = Sk + (k+1) [/tex]


    Now I'm worried that I didnt really solve anything. I'm totally knew at this, and I'm open to criticism and help, just be kind :)
    (ps.. this is my first time posting formulas, hopefully I did it right)


    Thanks
    dleacock
     
  2. jcsd
  3. Dec 15, 2005 #2
    just a repost to clean up the math...


    [tex] \frac{1}{4}(k+1)^2((k+1)+1)^2 + (k+1)^2 [/tex]

    [tex] = \sqrt{\frac{(k+1)^2((k+1)+1)^2}{4} + (k+1)^2} [/tex]

    [tex] = \frac{(k+1)((k+1)+1)}{2} + (k+1) [/tex]
     
  4. Dec 15, 2005 #3
    I don't own the book, but I looked at the pages on amazon, and it looks like you want to prove it using mathematical induction. This usually works in this way.

    Proof:
    Base Step (here, it will be n = 1)

    Inductive Step Assume it is true for some k, and then show that it is true for k+1.
    QED.


    Now you don't have to use induction, but it is what I think they want you to use. So first you should prove that the statement is true for n = 1, or k = 1. Meaning that if you plug 1 into the formula you will get the same thing, which is true.
    [tex]1^3 = (1)^2[/tex]
    [tex]=> 1 = 1[/tex]

    So then do the inductive hypothesis.

    You would say that the statement is true for some k, then show that it is true for k + 1.

    This would mean that you would want to show that:

    [tex]1^3 + 2^3 + 3^3 ... + k^3 + (k+1)^3 = (1 + 2 + 3 + ... + k + (k+1) )^2 [/tex]
    Using the fact that the statement is true for k.

    Meaning, prove the previous statement with the fact that: [tex]1^3 + 2^3 + 3^3 ... + k^3 = (1 + 2 + 3 + ... + k)^2 [/tex]

    --------

    I am not sure what you were doing here. But you can't just take the square root of something out of nowhere, also [tex]\sqrt{ a^2 + b^2 } \neq a + b[/tex] except in a few cases, so that is not allowed.
     
    Last edited: Dec 15, 2005
  5. Dec 15, 2005 #4
    yeah, thats exactly what I thouhgt the problem was. I just squared it out of nowhere. oh well, back to work I go. thanks for the response and the tip!
     
  6. Dec 16, 2005 #5
    In other words show, [tex]k^{2}(k+1)^{2} - k^{2}(k-1)^{2} = 4k^{3}[/tex]
     
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