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Question

Let α and β be roots of the equation x2x+1=0. Then the value of α+β+α2+β2++α100+β100 is

A
3
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B
1
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C
0
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D
3
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Solution

The correct option is A 3
x2x+1=0
So, α=ω and β=ω2, where ω is the cube root of unity.

Now, α+β+α2+β2++α100+β100
=(α+α2++α100)+(β+β2++β100)
=α(1α100)1α+β(1β100)1β
=ω(1ω100)1+ω+ω2(1ω200)1+ω2
=ωω2(1ω)+ω2ω(1ω2)
=1ω(1ω)+ω(1ω2)
=ω2(1ω)+ω(1ω2)
=ω21+ω1
=ω+ω22
=12
=3

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