What is the value of $ \sum _{k=1}^n\frac{x_k}{x_k-1}$? 
What is the value of 
  $$\sum _{k=1}^n\frac{x_k}{x_k-1}$$
  given that $x_1,x_2,\dots,x_n $ are the roots of the equation
  $x^n-3x^{n-1}+2x+1=0\,$? 

I wrote it as $ n+\sum _{k=1}^n\frac{1}{x_k-1}$ but didn't really help much.
I think Vieta's formulas come in handy here,
but I didn't get to the part where I could actually use any.
Could I have some hints on how to get this done?
Thank you. The answer is $S=3n-5$.
 A: We have
$$P(x)= x^n-3x^{n-1}+2x+1=\prod\limits_{k=1}^{n}\left(x-x_k\right)$$
From which
$$P'(x)=nx^{n-1}-(n-1)3x^{n-2}+2=\sum\limits_{i=1}^{n}\prod\limits_{k=1,k\ne i}^{n}\left(x-x_k\right)$$
then 
$$\frac{P'(x)}{P(x)}=\sum\limits_{i=1}^{n}\frac{1}{x-x_i}$$
But
$$S=n+\sum\limits_{k=1}^{n}\frac{1}{x_k-1}=n-\frac{P'(1)}{P(1)}=n-(n-3(n-1)+2)=3n-5$$
A: Let $p(x)=x^{n}-3x^{n-1}+2x+1=\prod_{i=1}^{n}(x-x_i)$. Then $$\frac{p'(x)}{p(x)}=\frac{\sum_{i=1}^{n}\prod_{1\leq j\leq n, j\neq i}(x-x_i)}{\prod_{i=1}^{n}(x-x_i)}=\sum_{i=1}^{n}\frac{1}{x-x_i}$$
Therefore $$\frac{xp'(x)}{p(x)}=\sum_{i=1}^{n}\frac{x}{x-x_i}=\sum_{i=1}^{n}\left(1+\frac{x_i}{x-x_i}\right)=n+\sum_{i=1}^{n}\frac{x_i}{x-x_i}$$
Evaluate at $x=1$, which we can because $p(1)=1\neq0$, to get 
$$\sum_{i=1}^{n}\frac{x_i}{x_i-1}=n-\frac{p'(1)}{p(1)}=3n-5$$
A: We have for a polynomial $P(x)$ of degree $n$ with roots $\alpha_1, \alpha_2, \ldots \alpha_n$,
$\frac{P'(a)}{P(a)} = \frac{1}{a-\alpha_1}+\frac{1}{a-\alpha_1}+\cdots \frac{1}{a-\alpha_n}$
Hence $\sum _{k=1}^n\frac{1}{x_k-1} = -\frac{P'(1)}{P(1)} = 2n-5$ and hence the sum of the given series is $3n-5$
A: Hint. You are on the right track. Let $x_k-1=1/y_k$ then
$$\sum _{k=1}^n\frac{x_k}{x_k-1}=n+\sum _{k=1}^n y_k.$$
For $x=1+1/y$ we have that
\begin{align*}0&=y^n(x^n-3x^{n-1}+2x+1)=
y^n((1/y+1)^n-3(1/y+1)^{n-1}+2(1/y+1)+1)\\
&=(y+1)^n-3y(y+1)^{n-1}+2(y^n+y^{n-1})+y^n\\
&=(1-3+2+1)y^n+(n-3(n-1)+2)y^{n-1}+o(y^{n-1})\\
&=y^{n}-(2n-5)y^{n-1}+o(y^{n-1}).
\end{align*}
Hence the sum of the roots, i.e. $\sum _{k=1}^n y_k$, is equal to $2n-5$ and we finally get
$$\sum _{k=1}^n\frac{x_k}{x_k-1}=n+(2n-5)=3n-5.$$
A: I think the $x_k$'s are the roots of the polynomial, right? 
Anyway, put $y_{k} = \frac{x_k}{x_k - 1}$, so that $x_k = \frac{y_k}{y_k - 1}$. Since $x_k$'s are the roots of the polynomial $f(x) = x^n - 3x^{n-1} + 2x + 1$, we have $f(y_k/(y_k-1)) =0$ for all $k$. Using this, one can find the polynomial that has $y_1, \dots, y_k$ as roots, and Vieta's formula will give the answer. 
A: One approach is to put $y=x-1$ so the sum becomes terms of the form $\cfrac {y+1}y=1+\cfrac 1y$
Now $y$ satisfies $$(y+1)^n-3(y+1)^{n-1}+2(y+1)+1=0$$
If we now divide through by $y^n$ and set $z=\cfrac 1y$ (it is possible to omit this step if you know what you are doing, and read off the relevant data from the polynomial for $y$) we get $$(1+z)^n-3z(1+z)^{n-1}+2z^{n-1}(1+z)+z^n=0$$and you should have a form you can use to conclude.
