<s>[INST] Compute $\sum\frac1{2-A_k}$ for $(A_k)$ the $n$th roots of unity 
If $1,A_1,A_2,A_3....A_{n-1}$ are the $n^{th}$ roots of unity then prove that 
  $$\dfrac{1}{2-A_1} + \dfrac{1}{2-A_2}+\cdots+ \dfrac{1}{2-A_{n-1}} = \dfrac{2^{n-1}(n-2) + 1}{2^n-1}$$

What I did: I tried to use some of the following formulas: 
$$1+ A_1 +A_2+A_3+\cdots+A_{n-1} = 0$$
$$\dfrac{2^n - 1}{2-1} = (2 -A_1)(2-A_2)\cdots(2-A_{n-1})$$
and the fact that $|A_i| = 1$ for $i =1,2,3,\cdots,n-1$.
 [/INST] Two ingredients for this proof:


*

*The numbers $A_k$ for $1\leqslant k<n$ and $A_n=1$ are the roots of the polynomial $X^n-1$.

*If $Y=\frac1{2-X}$ then $X=\frac{2Y-1}Y$. 


Thus, the numbers $B_k=\frac1{2-A_k}$ for $1\leqslant k<n$ and $B_n=\frac1{2-A_n}=1$ are the solutions of the equation $$\left(\frac{2Y-1}Y\right)^n-1=0$$ that is, the roots of the polynomial $$(2Y-1)^n-Y^n$$
The highest degree terms of this polynomial are $$(2^n-1)Y^n-n2^{n-1}Y^{n-1}+\ldots$$ hence the sum of its roots is $$\sum_{k=1}^nB_k=1+\sum_{k=1}^{n-1}\frac1{2-A_k}=\frac{n2^{n-1}}{2^n-1}$$ from which the desired result follows.
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