Question

Equation $x^n-1=0,n>1,nϵN$, has roots $1,a_2,....a_n$.

The value of $\sum _{r=2}^n\frac{1}{2-a_r}$, is

• A. $\frac{2^{n-1}(n-2)+1}{2^n-1}$
• B. $\frac{2^n(n-2)+1}{2^n-1}$
• C. $\frac{2^{n-1}(n-1)-1}{2^n-1}$
• D. none of these

Answer 1

The correct option is A $\frac{2^{n-1}(n-2)+1}{2^n-1}$

Equation $x^n-1=0,n>1,nϵN$, has roots $1,a_2,\dots a_n$.
We have to find the value of $\sum _{r=2}^n\frac{1}{2-a_r}$
$x^n-1=(x-1)(x-a_2)\dots (x-a_n)\to${i}
Taking log on both the sides
$\log (x^n-1)=\log (x-1)+\log (x-a_2)+\cdots +\log (x-a_n)$
Differentiating both the sides, we get
$\frac{n{x}^{n-1}}{x^n-1}=\frac{1}{x-1}+\frac{1}{x-a_2}+\cdots +\frac{1}{x-a_n}\to${ii}
Putting $x=2$ in eqn {ii}
$\frac{n{2}^{n-1}}{2^n-1}=1+\frac{1}{2-a_2}+\cdots +\frac{1}{2-a_n}$
$\frac{1}{2-a_2}+\cdots +\frac{1}{2-a_n}=\frac{n{2}^{n-1}}{2^n-1}-1=\frac{n{2}^{n-1}-2^n+1}{2^n-1}$
$\frac{1}{2-a_2}+\cdots +\frac{1}{2-a_n}=\frac{2^{n-1}(n-2)+1}{2^n-1}$