Question

The equation $x^n=1,n>1,n\in N$ has roots $1,a_1,a_2,...,a_{n-1}$. Then which of the following are correct?

• A. $\prod _{r=1}^{n-1}(1-a_r)=n$
• B. $\prod _{r=1}^{n-1}(1-a_r)=\frac{n}{2}$
• C. $\sum _{r=1}^{n-1}\frac{1}{1-a_r}=\frac{n+1}{2}$
• D. $\sum _{r=1}^{n-1}\frac{1}{1-a_r}=\frac{n-1}{2}$

Answer 1

Answer: (A), (D)

The correct options are
A $\prod _{r=1}^{n-1}(1-a_r)=n$
D $\sum _{r=1}^{n-1}\frac{1}{1-a_r}=\frac{n-1}{2}$

$1,a_1,a_2,a-3,...,a_{n-1}$ are the roots of $x^n-1=0$
$\Rightarrow x^n-1=(x-1)(x-a_1)(x-a_2)...(x-a_{n-1})$
So we can write
$\underset{x\to 1}{lim}\frac{x^n-1}{x-1}$
$=\underset{x\to 1}{lim}\left[\right(x-a_1\left)\right(x-a_2\left)\right(x-a_3)...(x-a_{n-1}\left)\right]$
$\Rightarrow n=(1-a_1)(1-a_2)(1-a_3)...(1-a_{n-1})$

Hence, $\prod _{r=1}^{n-1}(1-a_r)=n$

Now, $\log (x^n-1)=\log (x-1)+\log (x-a_1)+\log (x-a_2)+\cdots +\log (x-a_{n-1})$
Differentiate with respect to $x,$ we get
$\frac{n{x}^{n-1}}{x^n-1}=\frac{1}{x-1}+\frac{1}{x-a_1}+\frac{1}{x-a_2}+...+\frac{1}{x-a_{n-1}}$

$\Rightarrow \frac{n{x}^{n-1}}{x^n-1}-\frac{1}{x-1}=\frac{1}{x-a_1}+\frac{1}{x-a_2}+...+\frac{1}{x-a_{n-1}}$

$\Rightarrow \frac{n{x}^{n-1}-1(1+x+x^2+...+x^{n-1})}{x^n-1}=\frac{1}{x-a_1}+\frac{1}{x-a_2}+...+\frac{1}{x-a_{n-1}}$

$\Rightarrow \underset{x\to 1}{lim}\frac{n{x}^{n-1}-1(1+x+x^2+...+x^{n-1})}{x^n-1}=\underset{x\to 1}{lim}(\frac{1}{x-a_1}+\frac{1}{x-a_2}+...+\frac{1}{x-a_{n-1}})$

$\Rightarrow \frac{n(n-1)-(1+2+...+(n-1\left)\right)}{n}=\frac{1}{1-a_1}+\frac{1}{1-a_2}+\frac{1}{1-a_3}+...+\frac{1}{1-a_{n-1}}$

$\Rightarrow \frac{n(n-1)-n(n-1)/2}{n}=\frac{1}{1-a_1}+\frac{1}{1-a_2}+\frac{1}{1-a_3}+...+\frac{1}{1-a_{n-1}}$

Hence, $\sum _{r=1}^{n-1}(\frac{1}{1-a_r})=\frac{n-1}{2}$