Equation xn-1-0- n - 1- n- N- has roots 1- a2- -an-The value of 1-a11-a2-1-an- is

The correct option is B n x^n 1=0, n gt; 1 Since 1,a, a _ 2 ,....... a _ n are the roots of x ^ n 1=0 then, x ^ n 1=( x 1 ) ( x ...

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Roots of Quadratic Equation

Equation xn...

Question

Equation $x^{n} - 1 = 0, n > 1, n ϵ N$ , has roots $1, a_{2}, . . . . a_{n}$ .
The value of $(1 - a_{1}) (1 - a_{2}) . . . . (1 - a_{n})$ , is

\frac{n^{2}}{2}

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n

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(- 1)^{n} n

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none of these

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Solution

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Similar questions

x^{n} - 1 = 0, n > 1, n ϵ N

1, a_{2}, . . . . a_{n}

n \sum r = 2 \frac{1}{2 - a_{r}}

x^{n} - 1 = 0, n > 1, n \in N,

1, a_{1}, a_{2}, \dots, a_{n - 1},

x^{n} - 1 = 0

1, a_{1}, a_{2} . . . . a_{n - 1}

(1 - a_{1}) (1 - a_{2}) (1 - a_{3}) . . . . . (1 - a_{n - 1})

a_{1}, a_{2}, . . . . a_{n - 1}

n

(1 - a_{1}) (1 - a_{2}) (1 - a_{n - 1}) . . . = n

1 + a_{1} + a_{2} + . . . . + a_{n - 1} = 0

\frac{1}{2 - a_{1}} + \frac{1}{2 - a_{2}} + . . . . \frac{1}{2 - a_{n - 1}} = \frac{(n - 2) 2^{n - 1} + 1}{2^{n} - 1}

\frac{a_{0}}{n + 1} + \frac{a_{1}}{n} + \frac{a_{2}}{n - 1} + \dots + \frac{a_{n - 1}}{2} + a_{n} = 0

a_{0} x^{n} + a_{1} x^{n - 1} + \dots + a_{n - 1} x + a_{n} = 0

(0, 1)

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