Determine the value of $\sum_{i=1}^{2019}\frac{x_i}{x_i-1}$ If $ x_1, x_2, ...., x_ {2019} $ are roots of $ P (x) = x ^ {2019} + 2019x-1 $, determine the value of$$\sum_{i=1}^{2019}\frac{x_i}{x_i-1}$$
Solution: By Vieta, $x_1x_2..x_{2019}=1$,$\sum{x_1x_2..x_{2018}=2019}$, and the rest is $0$. Thus $$\sum_{i=1}^{2019}\frac{x_i}{x_i-1}=\frac{2019x_1x_2..x_{2019}-2018\sum{x_1x_2..x_{2018}}+2017\sum{x_1x_2..x_{2017}}-..}{x_1x_2..x_{2019}-\sum{x_1x_2..x_{2018}} +\sum{x_1x_2..x_{2017}} - .. -1}=\frac{2019-2018(2019)}{1-2019-1}=2017$$
The only thing I understood was $ x_1x_2..x_ {2019} = 1 $. Does this product sum of $ x_1 $ up to $ x_ {2018} $ mean what?
 A: Maybe this makes it more clear.
Note that the sum is equal to $$\sum_{i=1}^{2019}\left(1-\frac{1}{1-x_i}\right)=2019-\sum_{i=1}^{2019}\frac{1}{1-x_i}$$
Now, $\frac{P'(x)}{P(x)}=\sum_{i=1}^{2019}\frac{1}{x-x_i}$. Therefore, the last sum is equal to $\frac{P'(1)}{P(1)}$.
So, that numerator that they have there is $2019P(1)-P'(1)$.
By the way, the sum written as $\sum x_1x_2\ldots x_{2018}=2019$ should rather be more clear that each term is the product of all $x_i$, but one of them is missing in each term. Maybe a way to write this that is more clear is $\sum_{i=1}^{2019}\prod_{1\leq j\leq 2019, j\neq i}x_j$.
A: $$\frac{x_i}{x_i-1}=1+\frac1{x_i-1}.$$
If $x_i$ is a root of $x^{2019}+2019x-1=0$, then $x_i-1$ is a root of $Q(x):=(x+1)^{2019}+2019(x+1)-1=0$. And $\dfrac1{x_i-1}$ are the roots of that polynomial with the coefficients taken in the reverse order.
Finally, by Vieta,
$$\sum\dfrac1{x_i-1}=-\frac{Q_{1}}{Q_0}=-\frac{2019+2019}{1+2019-1}=-2$$
and the requested sum is $2017$.
A: Let $\dfrac x{x-1}=y,x=\dfrac y{y-1}$
$$(y/(y-1))^{2019}+2019 y/(y-1)-1=0$$
$$y^{2019}+2019y(y-1)^{2018}-(y-1)^{2019}=0$$
$$y^{2019}(1+2019-1)-y^{2018}(2019\cdot2018-2019)+\cdots=0$$
Now apply Vieta's formula
