New Year Problem (4)

Since $a_1,a_2,a_3,\dots,a_{2020}$ are the roots of the polynomial $f(x)\implies f(x)= (x-a_1)(x-a_2)(x-a_3)\dots(x-a_{2020})$

$x^{2020}+x^{2019}+x^{2018}+\dots+x^2+x = 1009.5\implies (x-a_1)(x-a_2)(x-a_3)\dots(x-a_{2020})-1009.5=0$

Therefore, by logarithmic differentiation, we obtain $\frac{f’(x)}{f(x)} =\frac{1}{(x-a_1)}+\frac{1}{(x-a_2)} +\frac{1}{(x-a_3)} + \dots. +\frac{1}{(x-a_{2020})}\implies$

$\sum_{n=1}^{2020}\frac{1}{1-a_n} = \frac{f’(1)}{f(1)} = \frac{1}{(1-a_1)}+\frac{1}{(1-a_2)} +\frac{1}{(1-a_3)} + \dots +\frac{1}{(1-a_{2020})}\implies$

$\sum_{n=1}^{2020}\frac{1}{1-a_n}= \frac{1+2+3+\dots+2019+2020}{2020-1009.5}=\frac{2020\times 2021/2}{2020-1009.5}=\boxed{2020}$