Newton's Binomial: Simplify sum[n=0,m] [1/(n + 1) * m-choose-n]

yossa

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May 18, 2018
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Hello, I would be happy to help with the solution,



Simplify the amount:

. . . . .\(\displaystyle \displaystyle \sum_{n=0}^m\, \dfrac{1}{n\, +\, 1}\, {m \choose n}\)



Reached the dependent expression in M, which contains no sums.
thanks :)
 

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