Question

# If 1,z1,z2,z3,⋯zn−1 are n roots of unity, then the value of 13−z1+13−z2+⋯+13−zn−1 is equal to :

A
n3n13n112
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B
n3n13n1+12
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C
n3n13n11
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D
n3n13n1+1
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Solution

## The correct option is A n⋅3n−13n−1−12xn=1 ⇒ xn−1=0 xn−1=(x−1)(x−z1)(x−z2)⋯(x−zn−1) where 1,z2,⋯,zn−1 are the n roots Now, taking log both sides log(xn−1)=log(x−1)+log(x−z1)+log(x−z2)+⋯+log(x−zn−1) differentiating both side w.r.t x, we get nxn−1xn−1=1x−1+1x−z1+1x−z2+⋯+1x−zn−1 Putting x=3, we get 13−z1+13−z2+⋯+13−zn−1=n⋅3n−13n−1−12 Alternate Solution: 1,z1,z2,...,zn−1 are the roots of zn−1=0 ⇒(z−1)(z−z1)(z−z2)...(z−zn−1)=0 ⇒1+z+z2+...+zn−1=0 Replace 13−z=y⇒z=3y−1y 1+3y−1y+(3y−1y)2+...+(3y−1y)n−1=0 Multiply by yn−1 ⇒yn−1+yn−2(3y−1)+yn−3(3y−1)2+...+(3y−1)n−1=0 ⇒yn−1(1+3+32+...+3n−1)− yn−2(1+2⋅3+3⋅32+4⋅33+...+(n−1)3n−2) +...+(−1)n−1=0 We know that, sum of roots =−ba ∴13−z1+13−z2+13−z3+...+13−zn−1=1+2⋅3+3⋅32+4⋅33+...+(n−1)3n−21+3+32+...+3n−1 =n−1∑r=1r⋅3r−1n∑r=13r−1 Let S=n−1∑r=1r⋅3r−1 S=1+2⋅3+3⋅32+4⋅33+...+(n−1)3n−2 3S= 3+2⋅32+3⋅33+...+(n−2)3n−2+(n−1)3n−1 Subtracting above equations, we get −2S=3n−1−12−(n−1)3n−1 ⇒S=n⋅3n−12−3n−14 S′=n∑r=13r−1=3n−12 Required Answer= =n⋅3n−13n−1−12

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