已知定义在 $\mathbb R$ 上的奇函数 $f(x)$ 满足 $f(x)=1$，且对任意 $x<0$，均有 $f\left(\dfrac 1x\right)=xf\left(\dfrac 1{1-x}\right)$，$n$ 是不小于 $2$ 的正整数，则 $\displaystyle \sum\limits_{k=1}^{n-1}f\left(\dfrac 1k\right)f\left(\dfrac{1}{n-k}\right)=$ _____．
答案 $\dfrac{2^{n-2}}{(n-2)!}$．
解析 根据题意，当 $k\in\mathbb N^{\ast}$ 时，有\[f\left(\dfrac 1k\right)=-f\left(-\dfrac 1k\right)=k\cdot f\left(\dfrac1{k+1}\right)\implies f\left(\dfrac1{k+1}\right)=\dfrac 1k\cdot f\left(\dfrac1k\right),\]而 $f(1)=1$，于是 $f\left(\dfrac 1k\right)=\dfrac{1}{(k-1)!}$（$k\in\mathbb N^{\ast}$）．因此\[ \sum\limits_{k=1}^{n-1}f\left(\dfrac 1k\right)f\left(\dfrac{1}{n-k}\right)=\sum_{k=1}^{n-1}\dfrac{1}{(k-1)!\cdot (n-k-1)!}=\sum_{k=1}^{n-1}\dfrac{\dbinom {n-2}{k-1}}{(n-2)!}=\dfrac{2^{n-2}}{(n-2)!}.\]