Find the value of ∑r=1412−ar where ak(k=0,1,2,3,4,5) are fifth roots of unity.

Another proof:
${\displaystyle P\left(x\right)=x^5-1=pro{f}_{r=0}^4(x-{\alpha }^r)}$
Hence
${\displaystyle \frac{P^{\prime }\left(x\right)}{P\left(x\right)}=\frac{5x^4}{x^5-1}=\sum _{r=0}^4\frac{1}{x-{\alpha }^r}}$
Now put x=2 in this formula.

Since ${\displaystyle a_0,a_1,a_2,\dots ,a_4}$ are roots of the equation
${\displaystyle x^5-1=0}$
You can apply Transformation of Roots to find a equation whose roots are
${\displaystyle \frac{1}{2-{\alpha }_0},\frac{1}{2-{\alpha }_1},\dots \frac{1}{2-{\alpha }_4}}$
Let ${\displaystyle P\left(y\right)}$ represent the polynomial whose roots are ${\displaystyle \frac{1}{2-a_k}}$
${\displaystyle y=\frac{1}{2-a_k}=\frac{1}{2-x}\Rightarrow x=\frac{2y-1}{y}}$
Put in (1)
${\displaystyle {\left(\frac{2y-1}{y}\right)}^5-1=0\Rightarrow {(2y-1)}^5-y^5=0}$
Use Binomial Theorem to find coefficient of ${\displaystyle y^5}$ and ${\displaystyle y^4}$.You will get sum of the roots using Vieta's Formulas.

If $a^5=1,\frac{1}{2-\alpha }=c_0+c_1\alpha +...+c_4{\alpha }^4$ where $2c_0-c_4=1$ $2c_0-c_4=1$ $2c_1-c_0=0$ $2c_2-c_2=0$ $2c_3-c_2=0$ $2c_4-c_3=0$ The solution of this is $c_0=16/31,c_1=8/31,c=4/31,c_3=2/31,c_4=1/31$ Then $\sum _{r=1}^4\frac{1}{2{\alpha }^r}=\frac{16}{31}\sum _{r=1}^{4}1+\frac{8}{31}\sum _{r=1}^4{a}^r+\frac{4}{31}\sum _{r=1}^4{a}^{2r}+\frac{2}{31}\sum _{r=1}^4{a}^{3r}+\frac{1}{31}\sum _{r=1}^4{a}^{4r}$ $=\frac{16}{31}\cdot 4-\frac{8}{31}-\frac{4}{31}-\frac{2}{31}-\frac{1}{31}=\frac{49}{31}$