Evaluate the sum
(1 + 1/ω) + (1 + 1/ω²) + (2 + 1/ω) + (2 + 1/ω²) + (3 + 1/ω) + (3 + 1/ω²) +... + (n + 1/ω) + (n + 1/ω²),
where ω is an imaginary cube root of unity.

Question

Accepted Answer

None of these. Since 1/w = w^2 and 1/w^2 = w, we have 1/w + 1/w^2 = w + w^2 = -1. Each pair (k+1/w)+(k+1/w^2) = 2k - 1, so the sum over k=1..n is sum(2k-1) = n^2, which is not among the first three forms, hence None of these.

Evaluate the sum (1 + 1/ω) + (1 + 1/ω²) + (2 + 1/ω) + (2 + 1/ω²) + (3 + 1/ω) + (3 + 1/ω²) +... + (n + 1/ω) + (n + 1/ω²), where ω is an imaginary cube root of unity.

Solution

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