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 A particularly handy but non-obvious form holds for the reciprocal power: 
:<math>\frac{1}{(1-x)^r}=\sum_{k=0}^\infty {r+k-1 \choose r-1} x^k.</math>
 
 For a more extensive account of Newton's generalized binomial theorem, see [[binomial series]].
 
 <eod> <sod> 
 
 A particularly handy but non-obvious form holds for the reciprocal power: 
:<math>\frac{1}{(1-x)^r}=\sum_{k=0}^\infty {r+k-1 \choose r-1} x^k \equiv \sum_{k=0}^\infty {r+k-1 \choose k} x^k.</math>
 
 For a more extensive account of Newton's generalized binomial theorem, see [[binomial series]].
 
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