Question

If ${\alpha }_0,{\alpha }_1,{\alpha }_2\dots ,{\alpha }_{n-1}$ be the n, nth roots of the unity, then the value of ${\sum }_{i=0}^{n-1}\frac{{\alpha }_i}{(3-{\alpha }_i)}$ is equal to

• A. $\frac{n}{3^n-1}$
• B. $\frac{n-1}{3^n-1}$
• C. $\frac{n+1}{3^n-1}$
• D. $\frac{n+2}{3^n-1}$

Answer 1

The correct option is A

$\frac{n}{3^n-1}$

Let $P={\sum }_{i=0}^{n-1}\frac{{\alpha }_i}{(3-{\alpha }_i)}=-{\sum }_{i=0}^{n-1}\frac{(3-{\alpha }_i)-3}{3-{\alpha }_i}=3{\sum }_{i=0}^{n-1}\frac{1}{3-{\alpha }_i}-{\sum }_{i=0}^{n-1}1\dots \dots \left(i\right)\because z^n-1={\sum }_{i=0}^{n-1}(z-{\alpha }_i)\therefore \ln (z^n-1)={\sum }_{i=0}^{n-1}\ln (z-{\alpha }_i)$
Differentiating both sides w.r.t. z, then
$\frac{n{z}^{n-1}}{(z^n-1)}={\sum }_{i=0}^{n-1}\frac{1}{(z-{\alpha }_i)}$
Using this relation to simplify Eq. (i), we get
$3\left(\frac{n{.3}^{n-1}}{3^n-1}\right)-n=\frac{n{3}^n}{3^n-1}-n=\frac{n}{3^n-1}$