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prove or disprove :

$$S=\dfrac{1}{1999}\binom{1999}{0}-\dfrac{1}{1998}\binom{1998}{1}+\dfrac{1}{1997}\binom{1997}{2}-\dfrac{1}{1996}\binom{1996}{3}+\cdots-\dfrac{1}{1000}\binom{1000}{999}=\dfrac{1}{1999}\left(w^{1999}_{1}+w^{1999}_{2}\right)$$ where $$w_{1}=\dfrac{1+\sqrt{3}i}{2},w_{2}=\dfrac{1-\sqrt{3}i}{2}$$

My idea: since $$\dfrac{1}{1999-k}\binom{1999-k}{k}=\dfrac{1}{1999-k}\dfrac{(1999-k)!}{k!(1999-2k)!}=\dfrac{(1998-k)!}{k!(1999-2k)!}$$

Then I can't,maybe can use integral deal it Thank you

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    $\begingroup$ the generating function is $-\log (1-x+x^2) = -\log (x-\omega_1)(x-\omega_2)$,.. $S_n = \dfrac{\omega_1^n+\omega_2^{n}}{n}$ $\endgroup$ – r9m May 23 '14 at 14:13
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The Chebyshev polynomials of the first kind are given by \begin{align} T_{n}(x) &= \frac{n}{2} \sum_{k=0}^{[n/2]} \frac{(-1)^{k}}{n-k} \binom{n-k}{k} (2x)^{n-2k} \\ &= \frac{1}{2} \left[ (x - \sqrt{x^{2}-1})^{n} + (x - \sqrt{x^{2}-1})^{n} \right]. \end{align} When $x=1/2$ it is seen that \begin{align} \sum_{k=0}^{[n/2]} \frac{(-1)^{k}}{n-k} \binom{n-k}{k} = \frac{1}{n} \left( a^{n} + b^{n} \right) \end{align} where $2a = 1+\sqrt{3} i$ and $2b=1-\sqrt{3} i$. When $n=1999$ this becomes \begin{align} \sum_{k=0}^{999} \frac{(-1)^{k}}{1999-k} \binom{1999-k}{k} = \frac{1}{1999} \left( a^{n} + b^{n} \right). \end{align} Thus the relationship is shown to be true.

If the sum is changed to all positive terms it can easily be seen as a Lucas number, namely, \begin{align} \sum_{k=0}^{999} \frac{1}{1999-k} \binom{1999-k}{k} = \frac{L_{1999}}{1999}. \end{align}

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This sum can also be done using Wilf / generatingfunctionology. Let $$a_n = \sum_{k=0}^{\lfloor n/2 \rfloor} \frac{(-1)^k}{n-k} {n-k \choose k}$$ and introduce the generating function $$f(z) = \sum_{n\ge 1} a_n z^n = \sum_{n\ge 1} \sum_{k=0}^{\lfloor n/2 \rfloor} \frac{(-1)^k}{n-k} {n-k \choose k} z^n \\ = \sum_{n\ge 1} \frac{z^n}{n} + \sum_{n\ge 1} \sum_{k=1}^{\lfloor n/2 \rfloor} \frac{(-1)^k}{n-k} {n-k \choose k} z^n.$$ Switch the order of summation in the inner sum to obtain $$\sum_{k\ge 1} \sum_{n\ge 2k} \frac{(-1)^k}{n-k} {n-k \choose k} z^n = \sum_{k\ge 1} \sum_{n\ge 0} \frac{(-1)^k}{n+k} {n+k \choose k} z^{n+2k} \\= \sum_{k\ge 1} z^{2k} \sum_{n\ge 0} \frac{(-1)^k}{n+k} {n+k \choose k} z^n = \sum_{k\ge 1} z^{2k} \sum_{n\ge 0} \frac{(-1)^k}{k} {n+k-1 \choose k-1} z^n \\ = \sum_{k\ge 1} \frac{(-1)^k}{k} z^{2k} \sum_{n\ge 0} {n+k-1 \choose k-1} z^n = \sum_{k\ge 1} \frac{(-1)^k}{k} z^{2k} \frac{1}{(1-z)^k} \\ = \log\frac{1}{1 + \frac{z^2}{1-z}}.$$ This gives the following closed form for $f(z):$ $$f(z) = \log\frac{1}{1-z} + \log\frac{1}{1 + \frac{z^2}{1-z}} = \log\frac{1}{1-z+z^2}.$$ Finally put $$w_{1,2} = \frac{1\pm i\sqrt{3}}{2}$$ so that $$1-z+z^2 = (1-w_1 z)(1-w_2 z).$$ This yields $$f(z) = \log\frac{1}{1-z+z^2} = \log\frac{1}{1-w_1 z} + \log\frac{1}{1-w_2 z}$$ so that $$[z^n] f(z) = \frac{w_1^n}{n} + \frac{w_2^n}{n},$$ as obtained by @r9m in the comment.

Remark. Here we have made repeated use of the identity $$\log\frac{1}{1-z} = \sum_{n\ge 1} \frac{z^n}{n}.$$

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