已知 $f(x)$ 是 $\mathbb R$ 上的奇函数，$f(1)=1$，且对任意 $x<0$，均有 $f\left(\dfrac{x}{x-1}\right)=xf(x)$．求 $f(1)f\left(\dfrac{1}{100}\right)+f\left(\dfrac{1}{2}\right)f\left(\dfrac{1}{99}\right)+f\left(\dfrac{1}{3}\right)f\left(\dfrac{1}{98}\right)+\cdots+f\left(\dfrac{1}{50}\right)f\left(\dfrac{1}{51}\right)$ 的值．
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【出处】
无
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【答案】
$\dfrac{2^{98}}{99!}$
【解析】
设 $a_{n}=f\left(\dfrac{1}{n}\right),n=1,2,3,\cdots$，则 $a_{1}=f(1)=1$．在 $f\left(\dfrac{x}{x-1}\right)=xf(x)$ 中取 $x=-\dfrac{1}{k}(k\in\mathbb N^{*})$，注意到\[\dfrac{x}{x-1}=\dfrac{-\dfrac{1}{k}}{-\dfrac{1}{k}-1}=\dfrac{1}{k+1},\]及 $f(x)$ 为奇函数，可知\[f\left(\dfrac{1}{k+1}\right)=-\dfrac{1}{k}\cdot f\left(-\dfrac{1}{k}\right)=\dfrac{1}{k}\cdot f\left(\dfrac{1}{k}\right),\]即 $\dfrac{a_{k+1}}{a_{k}}=\dfrac{1}{k}$，从而\[a_{n}=a_{1}\cdot \prod\limits_{k=1}^{n-1}\dfrac{a_{k+1}}{a_{k}}=\prod\limits _{k=1}^{n-1}\dfrac{1}{k}=\dfrac{1}{(n-1)!}.\]因此\[\begin{split}\sum\limits_{i=1}^{50}a_{i}a_{101-i}&=\sum\limits_{i=1}^{50}\dfrac{1}{(i-1)!(100-i)!}\\&=\sum\limits_{i=0}^{49}\dfrac{1}{i!(99-i)!}\\&=\dfrac{1}{99!}\cdot \sum\limits_{i=0}^{49}{\rm C}_{99}^{i}\\&=\dfrac{1}{99!}\cdot \sum\limits_{i=0}^{49}\dfrac{1}{2}({\rm C}_{99}^{i}+{\rm C}_{99}^{99-I})\\&=\dfrac{1}{99!}\cdot \dfrac{1}{2}\cdot 2^{99}=\dfrac{2^{98}}{99!}.\end{split}\]
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