If roots of an equation xn−1=0 are 1,a1,a2,…,an−1, then the value of 1−a11−a21−a3............(1-an-1) will be
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a
n
b
n2
c
nn
d
None of these
answer is A.
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Detailed Solution
Clearly,xn−1=(x−1)x−a1x−a2…x−an−1⇒ xn−1x−1=x−a1x−a2…x−anLHS is the summation of series in GP⇒ 1+x+x2+…+xn−1=x−a1x−a2…x−an−1⇒n=1−a11−a2…1−an−1 [putting x=1]