Summation over roots of unity Find the value of $\displaystyle\sum_{r=1}^{4} \frac{1}{2-\alpha^r} $
where $ \alpha^k (k=0,1,2,3,4,5) $ are fifth roots of unity.
My approach:- 
As we know that $ \alpha^k (k=0,1,2,3,4,5) $ are fifth roots of unity, then $ \alpha^k - 1$ should be equal to zero. Therefore the final answer to the summation $\displaystyle\sum_{r=1}^{4} \frac{1}{2-\alpha^r} $ should be $\displaystyle\sum_{r=1}^{4}  1 = 4  $
But the answer given is $ \dfrac{ 49}{31} $ 
Any help or hint will be much appreciated! 

If $\alpha^5 = 1$, $\frac{1}{2-\alpha} = c_0 + c_1 \alpha + \ldots + c_4 \alpha^4$ where 
$$\eqalign{ 2 c_0 - c_4 &= 1\cr
            2 c_1 - c_0 &= 0\cr
            2 c_2 - c_1 &= 0\cr
            2 c_3 - c_2 &= 0\cr
            2 c_4 - c_3 &= 0\cr}$$
The solution of this is $$c_0 = 16/31,\; c_1 = 8/31,\; c_2 = 4/31,\; c_3 = 2/31,\; c_4 = 1/31$$
Then $$ \eqalign{\sum_{r=1}^4 \frac{1}{2-\alpha^r} &= \frac{16}{31}\sum_{r=1}^4 1 + 
\frac{8}{31}\sum_{r=1}^4 \alpha^r + \frac{4}{31}\sum_{r=1}^4 \alpha^{2r} +
\frac{2}{31}\sum_{r=1}^4 \alpha^{3r} + \frac{1}{31}\sum_{r=1}^4 \alpha^{4r}\cr &= \frac{16}{31} \cdot 4 - \frac{8}{31} - \frac{4}{31} - \frac{2}{31} - \frac{1}{31} = \frac{49}{31}}$$
