Question

If $1,d_1,d_2,d_3\cdots ,d_{n-1}$ be $n^{th}$ roots of unity then which among the following are true

• A. $\frac{1}{1-d_1}+\frac{1}{1-d_2}+\cdots \frac{1}{1-d_{n-1}}=\frac{n-1}{2}$
• B. $(2+d_1)(2+d_2)\cdots (2+d_{n-1})=\frac{2^n+1}{3},$if $n$ is odd
• C. $(2-d_1)(2-d_2)\cdots (2-d_{n-1})=2^n-1$
• D. $(2+d_1)(2+d_2)\cdots (2+d_{n-1})=\frac{2^n+1}{3},$if $n$ is even

Answer 1

Answer: (A), (B), (C)

The correct options are
A $\frac{1}{1-d_1}+\frac{1}{1-d_2}+\cdots \frac{1}{1-d_{n-1}}=\frac{n-1}{2}$
B $(2+d_1)(2+d_2)\cdots (2+d_{n-1})=\frac{2^n+1}{3},$if $n$ is odd
C $(2-d_1)(2-d_2)\cdots (2-d_{n-1})=2^n-1$

$(x^n-1)=(x-1)(x-d_1)(x-d_2)\cdots (x-d_{n-1})$
$\Rightarrow (x-d_1)(x-d_2)\cdots (x-d_{n-1})=\frac{x^n-1}{x-1}=1+x+x^2+\cdots x^{n-1}\cdots \left(i\right)$
Put $x=2$
$\Rightarrow (2-d_1)(2-d_2)\cdots (2-d_{n-1})=1+2+2^2+\cdots +2^{n-1}$
Sum of $n$ terms of G.P $=\frac{1(2^n-1)}{2-1}$
Put $x=-2$
$\Rightarrow (-1{)}^{n-1}(2+d_1\left)\right(2+d_2)\cdots (2+d_{n-1})=\frac{(1-(-2{)}^n)}{1+2}\Rightarrow (2+d_1\left)\right(2+d_2)\cdots (2+d_{n-1})=(-1{)}^{1-n}\frac{(1-(-2{)}^n)}{1+2}$

Now if $n$ is even then sum will be $=\frac{2^n-1}{3}$
If $n$ is odd then sum will be $=\frac{2^n+1}{3}$
Differentiating $\left(i\right),$ we have
$1+2x+3x^2+(n-1)x^{n-2}=(x-d_1)(x-d_2)\cdots (x-d_{n-1})[\frac{1}{x-d_1}+\frac{1}{x-d_2}+\cdots +\frac{1}{x-d_{n-1}}]$
Put $x=1$
$\Rightarrow 1+2+3+\cdots (n-1)=n\cdot [\frac{1}{1-d_1}+\frac{1}{1-d_2}+\cdots \frac{1}{1-d_{n-1}}]$
$\Rightarrow [\frac{1}{1-d_1}+\frac{1}{1-d_2}+\cdots \frac{1}{1-d_{n-1}}]=\frac{(n-1)\cdot n}{2\cdot n}=\frac{n-1}{2}$