Equation `x^(n)-1=0, ngt1, n in N," has roots "1,a_(1),a_(2),…,a_(n-1).`
The value of `sum_(r=1)^(n-1) (1)/(2-a_(r))` is

Options

A. `(2^(n-1)(n-2)+1)/(2^(n)-1)`
B. `(2^(n)(n-2)+1)/(2^(n)-1)`
C. `(2^(n-1)(n-1)-1)/(2^(n)-1)`
D. none of these

Explanation

`"From "(1),log (x^(n)-1)=log (x-1)+log(x-a_(1))+...+log(x-a_(n-1))`
Differentiating w.r.t. x, we get
`(nx^(n-1))/(x^(n)-1)=(1)/(x-1)+(1)/(x-a_(1))+(1)/(x-a_(2))+...+(1)/(x-a_(n-1))" (2)"`
Putting x = 2 in (2), we get
`(n2^(n-1))/(2^(n)-1)=1+(1)/(2-a_(1))+(1)/(2-a_(2))+...+(1)/(2-a_(n-1))`
`"or " (1)/(2-a_(1))+(1)/(2-a_(2))+...+(1)/(2-a_(n-1))=(n2^(n-1))/(2^(n)-1)-1`
`=(n2^(n-1))/(2^(n)-1)`
`(2^(n-1)(n-2)+1)/(2^(n)-1)`