I stumbled upon the following problem:
Let $x_1, x_2, ..., x_n$ be the roots of the polynomial $P_n(x)=x^{n}+x^{n-1}+...+x+1$.
Prove that: $$\frac{1}{1-x_1}+...+\frac{1}{1-x_n}=\frac{n}{2}$$
I first thought about using induction, but could not figure out how to "link" $P_n$ with $P_{n-1}$. I also tried doing the first cases ($n=1,2,3,4$) but although they are feasible by using Vieta's formulas they did not shed light on the general case.
What should I do?
Hint:
Rewrite your polynôme as: $p(x)=(x-x_1)(x-x_2)\cdots (x-x_n)$
Takes the log derivative to get $$\dfrac{p’(x)}{p(x)}=\dfrac1{x-x_1}+\dfrac{1}{x-x_2}+\cdots+\dfrac1{x-x_m}$$
$$P_n(x)=x^{n}+x^{n-1}+...+x+1$$
$$(x-1)P_n(x)=x^{n+1}-1$$ so all the roots of $P_n$ are the roots of 1 which are not equal to 1, $\omega^k$ for $k=1...n$ where $\omega=e^{i\pi/(n+1)}$. Now notice that $$ {1\over 1-\omega^k}+{1\over 1-\omega^{-k}} ={1-\omega^k+1-\omega^{-k}\over1-\omega^k-\omega^{-k}+1} =1 $$ where I've used $-k$ as an equivalent for $n+1-k$. If $n$ is even, all the pairs of terms in the sum can thus be cancelled out, leaving us with $n/2$. If $n$ is odd, pairing gives us $(n-1)/2$, and then there is one unpaired term $\omega^{(n+1)/2}$. This is actually just -1, which is a root of $x+1$ and $x^3+x^2+x+1$ and so on, giving us a final extra term of $1/(1+(-1)^2)=\frac12$, so again we get $(n-1)/2+1/2=n/2$ for the sum.