Question

${\alpha }_0,{\alpha }_1,{\alpha }_2,.......{\alpha }_{n-1}$ be the $n,n^{th}$ roots of the unity, then the value $\sum _{i=0}^{n-1}\frac{{\alpha }_i}{3-{\alpha }_i}$ 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 $x=1^{\frac{1}{n}}\Rightarrow x^n=1$

$\therefore x^n-1=0$

or $x^n-1=(x-{\alpha }_0)(x-{\alpha }_1)(x-{\alpha }_2)...(x-{\alpha }_{n-1})={\prod }_{i=0}^{n-1}(x-{\alpha }_i)$

On taking logarithm both sides, we get

${\log }_e(x^n-1)={\sum }_{i=0}^{n-1}{\log }_e(x-{\alpha }_i)$

On differentiating both sides w.r.t. $x$, we get

$\frac{n{x}^{n-1}}{x^n-1}={\sum }_{i=0}^{n-1}\left(\frac{1}{x-{\alpha }_i}\right)$

On putting $x=3$ we get

$\frac{n{3}^{n-1}}{3^n-1}={\sum }_{i=0}^{n-1}\left(\frac{1}{3-{\alpha }_i}\right)$

Now, ${\sum }_{i=0}^{n-1}\frac{{\alpha }_i}{3-{\alpha }_i}={\sum }_{i=0}^{n-1}(-1+\frac{3}{3-{\alpha }_i})$

$=-{\sum }_{i=0}^{n-1}1+3{\sum }_{i=0}^{n-1}\frac{1}{3-{\alpha }_i}$

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

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

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

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