Question

If ${\alpha }_0,{\alpha }_1,{\alpha }_2,.........{\alpha }_{n-1}$ be the n, $n^{th}$ roots of unity, then 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. $\frac{n+2}{3^n-1}$
• 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 ------------(i)

$Z^n$ - 1 = $\underset{i=0}{\stackrel{n-1}{\Pi }}$ $(Z-{\alpha }_i)$

then log($Z^n$ - 1) = $\sum _{i=0}^{n-1}$$ln(Z-{\alpha }_i)$

Diff. both sides w.r.t. Z

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

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

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