Solveeit Logo
Login

Question

Question: Equation $x^n - 1 = 0$, $n > 1$, $n \in N$ has roots $1, a_1, a_2, \dots, a_{n-1}$. The value of $(...

Equation xn−1=0x^n - 1 = 0, n>1n > 1, n∈Nn \in N has roots 1,a1,a2,…,an−11, a_1, a_2, \dots, a_{n-1}.

The value of (1−a1)(1−a2)…(1−an−1)(1-a_1)(1-a_2)\dots(1-a_{n-1}) is:

A

n22\frac{n^2}{2}

B

nn

C

(−1)n⋅n(-1)^n \cdot n

D

00

Answer

n

Explanation

Solution

To find the value of (1−a1)(1−a2)…(1−an−1)(1-a_1)(1-a_2)\dots(1-a_{n-1}), substitute x=1x=1 into the expression for Q(x)Q(x), where Q(x)=xn−1x−1=xn−1+xn−2+⋯+x+1Q(x) = \frac{x^n - 1}{x-1} = x^{n-1} + x^{n-2} + \dots + x + 1.

(1−a1)(1−a2)…(1−an−1)=Q(1)(1-a_1)(1-a_2)\dots(1-a_{n-1}) = Q(1) Q(1)=1n−1+1n−2+⋯+1+1Q(1) = 1^{n-1} + 1^{n-2} + \dots + 1 + 1

Since there are nn terms, each equal to 1, their sum is nn.

Therefore, (1−a1)(1−a2)…(1−an−1)=n(1-a_1)(1-a_2)\dots(1-a_{n-1}) = n.