Question

# If 1,d1,d2,d3⋯,dn−1 be nth roots of unity then which among the following are true

A
11d1+11d2+11dn1=n12
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B
(2+d1)(2+d2)(2+dn1)=2n+13,if n is odd
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C
(2d1)(2d2)(2dn1)=2n1
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D
(2+d1)(2+d2)(2+dn1)=2n+13,if n is even
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Solution

## The correct options are A 11−d1+11−d2+⋯11−dn−1=n−12 B (2+d1)(2+d2)⋯(2+dn−1)=2n+13,if n is odd C (2−d1)(2−d2)⋯(2−dn−1)=2n−1(xn−1)=(x−1)(x−d1)(x−d2)⋯(x−dn−1) ⇒(x−d1)(x−d2)⋯(x−dn−1)=xn−1x−1=1+x+x2+⋯xn−1⋯(i) Put x=2 ⇒(2−d1)(2−d2)⋯(2−dn−1)=1+2+22+⋯+2n−1 Sum of n terms of G.P =1(2n−1)2−1 Put x=−2 ⇒(−1)n−1(2+d1)(2+d2)⋯(2+dn−1)=(1−(−2)n)1+2⇒(2+d1)(2+d2)⋯(2+dn−1)=(−1)1−n(1−(−2)n)1+2 Now if n is even then sum will be =2n−13 If n is odd then sum will be =2n+13 Differentiating (i), we have 1+2x+3x2+(n−1)xn−2=(x−d1)(x−d2)⋯(x−dn−1)[1x−d1+1x−d2+⋯+1x−dn−1] Put x=1 ⇒1+2+3+⋯(n−1)=n⋅[11−d1+11−d2+⋯11−dn−1] ⇒[11−d1+11−d2+⋯11−dn−1]=(n−1)⋅n2⋅n=n−12

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