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Where can i find the proof to this result?

  1. Sep 6, 2009 #1
    If n and r are positve integers, then there are constants


    such that:

    [tex]1^r+2^r+3^r+...+n^r= a_1n+a_2n^2+a_3n^3+...+a_{r+1}n^{r+1}[/tex]

    So,as the title says, can anyone either show me how to prove this, or rederect me to some other source(book, webpage) where i could find such proof.

    This is not hw!

  2. jcsd
  3. Sep 6, 2009 #2
    As it stands, the proposition is false. It implies that the left-hand side is divisible by n. But [itex]1^3 + 2^3[/itex] is not divisible by 2, neither is [itex]1^3 + \ldots + 6^3[/itex] divisible by 6.

    The right-hand side looks very much like a base-n expansion, though, so if you simply forgot to include an [itex]a_0 < n[/itex] term, the proof might go:

    [itex]1^r + \ldots + n^r \leq (n+1)n^r \leq n^{r+2}[/itex], and every number lower than [itex]n^{r+2}[/itex] can be written in the desired form.
  4. Sep 6, 2009 #3
    Dear Preno,

    I thank you for your reply. I don't think the proposition is false. Your reasoning would be true if the coefficients [tex] a_i, i \in N[/tex] would be asked to be integers. But there is no such restriction on the coefficients, they can be any real number.

    In general, as you know, we say that an integer n is divisible by another integer m,iff there exists another integer r, such that n=rm. but here like i said the coefficients are not restricted.

    The source where i got this suggests that this can be proven using the theory of linear differential equations, but i see no such relation so far.

    Any suggestions?
  5. Sep 6, 2009 #4
    If the [tex]a_i[/tex] are real numbers, the proposition is trivial. Let [tex] a_2 = \cdots = a_{r + 1} = 0[/tex] and let [tex]a_1 = \left( 1^r + \cdots + n^r \right) / n[/tex], and the proposition holds.
  6. Sep 6, 2009 #5
    Bernouilli polynomials
  7. Sep 6, 2009 #6
    He said "constants" to rule out this case. He wants to generalize:

    1+2+...+n = n^2/2+n/2

    1^2+2^2+...+n^2 = n^3/3+n^2/2+n/6
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