填充6
不失一般性可設\(xf(x)=a(x-1)(x-2)\ldots(x-2018)+1\)
則以\(x=0\)代入\(0=a \times 2018!+1\)，\( \displaystyle a=-\frac{1}{2018!} \)
故\( \displaystyle 2020f(2020)=-\frac{1}{2018!}\times 2019!+1 \)
\( \displaystyle f(2020)=-\frac{2018}{2020}=-\frac{1009}{1010} \)
