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Question
Let α,β denote the cube roots of unity other than 1 and αβ. Let s=∑2n=0(−1)n(αβ)n. Then the value of s is
Let α,β denote the cube roots of unity other than 1 and αβ. Let s=∑2n=0(−1)n(αβ)n. Then the value of s is
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Solution
The correct option is B Either −2ω or −2ω2
Given α,β are cube root of unity i.e. α,β are ω,ω2S=∑2n=0(−1)n(αβ)nS=1+(−1)(αβ)+(−1)2(αβ)2S=1−(ωω2)+(ωω2)2,1−(ω2ω)+(ω2ω)2S=1−1ω+1ω2,1−ω+ω2S=ω2−ω+1ω2,1−ω+ω2S=ω3−ω2+ω1,1−ω+ω2S=1−ω2+ω,1−ω+ω2We know that, 1+ω+ω2=0S=−ω2−ω2,−ω−ωS=−2ω2,−2ω
Given α,β are cube root of unity i.e. α,β are ω,ω2
S=∑2n=0(−1)n(αβ)nS=1+(−1)(αβ)+(−1)2(αβ)2S=1−(ωω2)+(ωω2)2,1−(ω2ω)+(ω2ω)2S=1−1ω+1ω2,1−ω+ω2S=ω2−ω+1ω2,1−ω+ω2S=ω3−ω2+ω1,1−ω+ω2S=1−ω2+ω,1−ω+ω2
We know that, 1+ω+ω2=0S=−ω2−ω2,−ω−ωS=−2ω2,−2ω
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