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Sum of powers of integers

  1. Apr 6, 2006 #1
    Let,s suppose we want to do this sum:

    [tex] 1+2^{m}+3^{m}+............+n^{m} [/tex] n finite

    then we could use the property of the differences:

    [tex] \sum_{n=0}^{n}(y(k)-y(k-1))=y(n)-y(0) [/tex]

    so for any function of the form f(x)=x^{m} m integer you need to solve:

    [tex] y(n)-y(n-1)=n^{m} [/tex] i don,t know how to solve

    it..:frown: :frown: i have tried the ansatz y(n)=K(n) with K(n) a Polynomial of degree m+1 but i don,t get the usual results for the sum..could someone help?..
  2. jcsd
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