Summation over roots of unity

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.

Note that $$\newcommand{\al}{\alpha}\frac1{2-\al^r} =\sum_{k=0}^\infty\frac{\alpha^{kr}}{2^{k+1}}$$ and so $$\sum_{r=0}^4\frac1{2-\al^r} =\sum_{k=0}^\infty\frac{1}{2^{k+1}}\sum_{r=0}^4\al^{kr}.$$ The inner sum is zero, unless $5\mid r$. So $$\sum_{r=0}^4\frac1{2-\al^r}=5\sum_{s=0}^\infty\frac1{2^{5s+1}} =\frac{5}{2(1-1/32)}=\frac{80}{31}.$$ Now subtract $1/(2-1)=1$ to get $49/31$.

Since $\alpha_0,\alpha_1,\alpha_2 \dots \alpha_{4}$ are roots of the equation

$$x^5-1=0 \tag1$$

You can apply Transformation of Roots to find a equation whose roots are$$\frac{1}{2-\alpha_0} , \frac{1}{2-\alpha_1},\dots \frac{1}{2-\alpha_{4}}$$

Let $P(y)$ represent the polynomial whose roots are $\frac{1}{2-\alpha_k}$

$$y=\frac{1}{2-\alpha_k}=\frac{1}{2-x} \implies x=\frac{2y-1}{y}$$

Put in $(1)$

$$\Bigg(\frac{2y-1}{y}\Bigg)^5-1=0 \implies (2y-1)^{5}-y^{5}=0$$

Use Binomial Theorem to find coefficient of $y^5$ and $y^{4}$.You will get sum of the roots using Vieta's Formulas.

Hope it helps!