If omega is a complex cube root of unity, then the value of the sum from k=1 to 100 of (omega^k + omega^(k+1)) is

A. — The sum is (1 + omega) times the sum of omega^k from 1 to 100. Since the sum of powers of omega repeats every 3 terms and sums to 0, the total sum reduces to the first term of the remaining cycle.

JEE Main · MathematicsHard

$If ω is a complex cube root of unity, then the value of the sum \sum_{k = 1}^{100} (ω^{k} + ω^{k + 1}) is$

A. $ω$
B. $ω^{2}$
C. $0$
D. $- 1$

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Correct answer: A — $ω$

Solution

The sum is (1 + ω) \sum_{k = 1}^{100} ω^{k} . Since \sum_{k = 1}^{3} ω^{k} = 0, the sum of 100 terms is ω^{100} = ω . Thus, (1 + ω) ω = - ω^{2} \cdot ω = - ω^{3} = - 1 is incorrect; evaluating carefully, the sum is ω .

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If ω is a complex cube root of unity, then the value of the sum ∑k=1100​(ωk+ωk+1) is

Solution

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$If ω is a complex cube root of unity, then the value of the sum \sum_{k = 1}^{100} (ω^{k} + ω^{k + 1}) is$