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 $i =1,2,3,\cdots,n-1$.

Note that
$$
z^n-1=\prod_{k=0}^{n-1}(z-a_k)
$$
so that
$$
nz^{n-1}=\sum_{k=0}^{n-1}\frac1{z-a_k}\prod_{k=0}^{n-1}(z-a_k)
$$
and therefore,
$$
\frac{nz^{n-1}}{z^n-1}=\sum_{k=0}^{n-1}\frac1{z-a_k}
$$
Thus,
$$
\sum_{k=0}^{n-1}\frac1{2-a_k}=\frac{n2^{n-1}}{2^n-1}
$$
Subtracting the $k=0$ ($a_0=1$) term gives
$$
\sum_{k=1}^{n-1}\frac1{2-a_k}=\frac{(n-2)2^{n-1}+1}{2^n-1}
$$
