Dayton Burnett

2022-01-21

Find the value of $\sum _{r=1}^{4}\frac{1}{2-{a}^{r}}$ where ${a}^{k}\left(k=0,1,2,3,4,5\right)$ are fifth roots of unity.

cevarnamvu

Expert

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

eris0cg

Expert

Since ${a}_{0},{a}_{1},{a}_{2},\dots ,{a}_{4}$ are roots of the equation
${x}^{5}-1=0$
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 $P\left(y\right)$ represent the polynomial whose roots are $\frac{1}{2-{a}_{k}}$
$y=\frac{1}{2-{a}_{k}}=\frac{1}{2-x}⇒x=\frac{2y-1}{y}$
Put in (1)
${\left(\frac{2y-1}{y}\right)}^{5}-1=0⇒{\left(2y-1\right)}^{5}-{y}^{5}=0$
Use Binomial Theorem to find coefficient of ${y}^{5}$ and ${y}^{4}$.You will get sum of the roots using Vieta's Formulas.

RizerMix

Expert

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