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Binomial expansion?

  1. Dec 17, 2007 #1
    The next-to-last step in the proof on pg 1 of this article


    makes this substitution

    [tex] \sum_{r=0}^\infty \binom {m+r-2}{r} q^r = (1-q)^{1-m} [/tex]

    I don't see it. How does
    [tex] (1-q)^{1-m} = \sum_{r=0}^\infty \binom{1-m}{r} (-q)^r[/tex]
    [tex] (1-q)^{1-m} = \sum_{r=0}^\infty \binom{1-m}{r} (-q)^{1-m-r}[/tex]
    transform to the expression given by Haldane?
    Last edited by a moderator: Apr 23, 2017
  2. jcsd
  3. Dec 17, 2007 #2
    I just remembered that this m is always a positive integer so (1-m) is 0 or negative.

    The key to this must be here...

    http://mathworld.wolfram.com/NegativeBinomialSeries.html" [Broken]
    Last edited by a moderator: May 3, 2017
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