If 1,a1,a2...,anā1 are nth roots of unity, then 11āa1+11āa2+...+11āanā1 equals
A
2n−1n
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B
n−12
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C
nn−1
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D
None of these
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Solution
The correct option is Bn−12 Given zn=1,z=1,a1,a2,...,an−1 Let a=11−z⇒z=1−1a ∴(1−1a)n=1 ⇒(a−1)n−an=0 ⇒−C1an−1+C2an−2+...+(−1)n=0 where a=11−a1,11−a2.....11−an−1 ⇒11−a1+11−a2+.....+11−an−1=nc2n=n−12