每日一题[986]组合数求和

设$S=\dfrac{1}{2017}{\rm C}_{2017}^0-\dfrac{1}{2016}{\rm C}_{2016}^1+\dfrac{1}{2015}{\rm C}_{2015}^2-\cdots-\dfrac{1}{1010}{\rm C}_{1010}^{1007}+\dfrac{1}{1009}{\rm C}_{1009}^{1008}$，则$S$的值是_______．

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正确答案是$\dfrac{1}{2017}$．

分析与解　根据题意，有

$$\begin{split}2017S&={\rm C}_{2017}^0-\left(1+\dfrac 1{2016}\right){\rm C}_{2016}^1+\left(1+\dfrac{2}{2015}\right){\rm C}_{2015}^2-\cdots+\left(1+\dfrac{1008}{1009}\right){\rm C}_{1009}^{1008}\\&={\rm C}_{2017}^0-\left({\rm C}_{2016}^1+{\rm C}_{2015}^0\right)+\left({\rm C}_{2015}^2+{\rm C}_{2014}^1\right)-\cdots+\left({\rm C}_{1009}^{1008}+{\rm C}_{1008}^{1007}\right)\\&=S_{2017}-S_{2015},\end{split}$$ 其中 $$\begin{split}S_{2017}&={\rm C}_{2017}^0-{\rm C}_{2016}^1+{\rm C}_{2015}^2-\cdots+{\rm C}_{1009}^{1008},\\ S_{2015}&={\rm C}_{2015}^0-{\rm C}_{2014}^1+{\rm C}_{2013}^2-\cdots-{\rm C}_{1008}^{1007},\end{split}$$ 类似的定义 $$\begin{split}S_{2k}&={\rm C}_{2k}^0-{\rm C}_{2k-1}^1+{\rm C}_{2k-2}^2-\cdots+(-1)^k{\rm C}_k^k,\\S_{2k+1}&={\rm C}_{2k+1}^0-{\rm C}_{2k}^1+{\rm C}_{2k-1}^2-\cdots+(-1)^k{\rm C}_{k+1}^k,\end{split}$$ 那么由组合恒等式 $${\rm C}_n^m={\rm C}_{n-1}^{m-1}+{\rm C}_{n-1}^m,$$ 可得 $$S_{2k+1}-S_{2k}=-S_{2k-1},S_{2k+2}-S_{2k+1}=-S_{2k},$$ 因此有 $$S_{n+2}=S_{n+1}-S_n,$$ 于是数列$\{S_n\}$是周期为$6$的数列．考虑到$S_1=1$，$S_2=0$，因此 $$S_n:\underbrace{1,0,-1,-1,0,1,1},\underbrace{1,0,-1,-1,0,1,1},\cdots,$$ 因此 $$S_{2017}=S_1=1,S_{2015}=S_5=0,$$ 进而所求代数式的值为$\dfrac{1}{2017}$．