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rational_numbers [2016/04/23 13:41]
nikolaj
rational_numbers [2016/04/23 13:49]
nikolaj
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 For all $m$ For all $m$
  
-$\dfrac{1}{x}=1 + \sum_{k=1}^m (1-x)^k \dfrac{1}{x} (1-x)^{m+1}$+$\sum_{k=0}^m x^k = \dfrac{1}{1-x}(1-x^{m+1})$ ^ 
 +^ $\sum_{k=0}^m (1-y)^k \dfrac{1}{y}-\dfrac{1}{y}(1-y)^{m+1}$ ​^
  
 == Logic == == Logic ==
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