<s>[INST] Series involving complex roots: $\frac{1}{2-a_1} + \frac{1}{2-a_2} + \dots + \frac{1}{2-a_{n-1}} = \frac{(n-2)2^{n-1}+1}{2^n - 1}$ $$
\frac{1}{2-a_1} + \frac{1}{2-a_2} + \dots + \frac{1}{2-a_{n-1}} = 
\frac{(n-2)2^{n-1}+1}{2^n - 1}
$$
Here $1,a_1,a_2,\dots,a_{n-1}$ are $n$-th roots of unity
I know the sum of roots is 0. I think the series should be some sort of telescopic.
 [/INST] Let $y=\dfrac1{2-x}\iff x=\dfrac{2y-1}y$ where $x^n=1$
$\implies\left(\dfrac{2y-1}y\right)^n=1$
$\displaystyle\iff y^n(2^n-1)-\binom n1 2^{n-1}y^{n-1}+\cdots+(-1)^n=0$
Using Vieta's formula, $$\displaystyle\sum_{r=0}^{n-1}y_r=\dfrac{\binom n1 2^{n-1}}{2^n-1}=\dfrac{n 2^{n-1}}{2^n-1}$$
If $x=1,y=1=y_0$(say)
$\displaystyle\implies\sum_{r=1}^{n-1}y_r=\dfrac{n 2^{n-1}}{2^n-1}-y_0=\dfrac{n 2^{n-1}}{2^n-1}-1=\cdots$
</s>