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!
Best Answer
Another proof: $$P(x)=x^5-1=\prod_{r=0}^4(x-\alpha^r)$$ Hence $$\frac{P^{\prime}(x)}{P(x)}=\frac{5x^4}{x^5-1}=\sum_{r=0}^4\frac{1}{x-\alpha^r}$$ Now put $x=2$ in this formula.