If x 6 = 2 x 3 − 1 and x is nonreal, evaluate ∑ r = 1 50 ( x r + x 2 r ) 3 .

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MathematicsSequence & SeriesSpecial sequences/seriesMedium2 minQB

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If $x^{6} = 2 x^{3} - 1$ and $x$ is nonreal, evaluate $\sum_{r = 1}^{50} (x^{r} + x^{2 r})^{3}$ .

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Solution:

$(x^{3} - 1)^{2} = 0$ , so $x$ is a nonreal cube root of unity. Then $x^{r} + x^{2 r} = 2$ when $3 ∣ r$ (16 cases) and -1 otherwise (34 cases). The sum of cubes is $16 (8) - 34 = 94$ .

Stream:JEESubject:MathematicsTopic:Sequence & SeriesSubtopic:Special sequences/series

⏱ 2mℹ️ Source: QB

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