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The next-to-last step in the proof on pg 1 of this article

http://links.jstor.org/sici?sici=0006-3444(194511)33:3<222:OAMOEF>2.0.CO;2-Whttp://links.jstor.org/sici?sici=0006-3444%28194511%2933%3A3%3C222%3AOAMOEF%3E2.0.CO%3B2-W

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]

or

[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?

http://links.jstor.org/sici?sici=0006-3444(194511)33:3<222:OAMOEF>2.0.CO;2-Whttp://links.jstor.org/sici?sici=0006-3444%28194511%2933%3A3%3C222%3AOAMOEF%3E2.0.CO%3B2-W

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]

or

[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?

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