Question

If ${\alpha }_0,{\alpha }_1,{\alpha }_2,\dots \dots \dots \dots \dots {\alpha }_{n-1}$ be the n${}^{th}$ roots of unity, then the value of $\begin{array}{c}n-1\\ \Sigma \\ i=0\end{array}\frac{{\alpha }_i}{3-{\alpha }_i}$ is equal to

• A.
• B.
• C.
• D. None of these

Answer 1

The correct option is A

Since ${\alpha }_0,{\alpha }_1,{\alpha }_2\dots \dots \dots \dots {\alpha }_{n-1}are{n}^{th}$ roots of unity.
$\therefore x^n-1=(x-{\alpha }_0)(x-{\alpha }_1)\dots \dots \dots ..(x-{\alpha }_{n-1})$
$⟹\log (x^n-1)=\log (x-{\alpha }_0)+\log (x-{\alpha }_1)+\dots \dots +\log (x-{\alpha }_{n-1})$
On differentiating both sides wrt x, we get $\frac{{nx}^{n-1}}{x^n-1}=\frac{1}{x-{\alpha }_0}+\frac{1}{x-{\alpha }_1}+\dots \dots \dots ..+\frac{1}{x-{\alpha }_{n-1}}$ on putting x = 3 on both sides, we get
$\frac{{n3}^{n-1}}{3^n-1}$ $\frac{1}{x-{\alpha }_0}=\frac{1}{3-{\alpha }_1}+\frac{1}{3-{\alpha }_{n-1}}+\dots \dots \dots \dots +\frac{1}{3-{\alpha }_{n-1}}\dots \dots \dots \dots \dots .\left(i\right)$
Now, $\begin{array}{c}n-1\\ \Sigma \\ i=0\end{array}\frac{{\alpha }_i}{3-{\alpha }_i}=-\begin{array}{c}n-1\\ \Sigma \\ i=0\end{array}\frac{\{(3-{\alpha }_i)-3\}}{(3-{\alpha }_i)}$= $-\begin{array}{c}n-1\\ \Sigma \\ i=0\end{array}1+3\begin{array}{c}n-1\\ \Sigma \\ i=0\end{array}\frac{1}{3-{\alpha }_i}$
= $-$ n + 3 $\times \frac{{n3}^{n-1}}{3^n-1}\left[usingeq\right(i\left)\right]$
= $-$ n + n $\frac{{n3}^{n-1}}{3^n-1}$ = $\frac{n}{3^n-1}$