Show using antiderivatives and using the binomial coefficient.
Show ${n \choose 0} - \frac{1}{2} {n \choose 1} + ... + \frac{(-1)^n}{n+1} {n \choose n} = \frac{1}{n+1}$
-1
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combinatorics
number-theory
1 Answers
1
$$f(x)=(x-1)^n=\sum_{k=0}^n{n\choose k}(-1)^{n-k}x^k$$ then $$\int_0^x f(x)=\frac{(x-1)^{n+1}-(-1)^{n+1}}{n+1}=\sum_{k=0}^n{n\choose k}(-1)^{n-k}\frac{x^{k+1}}{k+1}$$ now $x=1$.
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0how are you getting $\frac{(x-1)^{n+1}\textbf{-1}}{n+1}$? where does the -1 come from? – 2017-02-15