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Question: If C<sub>r</sub> = \(\left( \begin{aligned} & n \\ & r \end{aligned} \right)\), then \(\sum_{r = 1...

If Cr = (nr)\left( \begin{aligned} & n \\ & r \end{aligned} \right), then ∑r=1n(−1)r−1Crr\sum_{r = 1}^{n}{( - 1)^{r - 1}\frac{C_{r}}{r}} =

A

0

B

1

C

∑r=1n1r\sum_{r = 1}^{n}\frac{1}{r}

D

∑r=1n(−1)r−1r\sum_{r = 1}^{n}\frac{( - 1)^{r - 1}}{r}

Answer

∑r=1n1r\sum_{r = 1}^{n}\frac{1}{r}

Explanation

Solution

1−(1−x)nx\frac{1 - (1 - x)^{n}}{x} = ∑r=1n(−1)n−1Crxr−1\sum_{r = 1}^{n}{( - 1)^{n - 1}C_{r}x^{r - 1}}

Integrating between 0 and 1 yields the desired sum ∑r=1n1r\sum_{r = 1}^{n}\frac{1}{r}