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BITSAT 2015
Exam
BITSAT
Year
2015
Subject
Mathematics
Chapter
Sequences and Series

BITSAT 2015 · Mathematics · Sequences and Series

If ω is the complex cube root of unity, then the value of ω+ω^...

If ωisthecomplexcuberootofunity,thenthevalueof\omega is the complex cube root of unity, then the value of\omega+\omega^{\left(\frac{1}{2}+\frac{3}{8}+\frac{9}{32}+\frac{27}{128}+\ldots\right)} is

Options

  1. A-1
  2. B1
  3. C−i-\mathrm{i}
  4. Di

Correct answer

A. -1

Step-by-step solution

Consider 12+38+932+27128+….\frac{1}{2}+\frac{3}{8}+\frac{9}{32}+\frac{27}{128}+\ldots . Which can be written as \frac{3^{0}}{2^{1}}+\frac{3^{1}}{2^{3}}+\frac{3^{2}}{2^{5}}+\frac{3^{3}}{2^{7}}+\ldots .=\frac{1}{2} \left[1+\frac{3}{2^{2}}+\frac{3^{2}}{2^{4}}+\frac{3^{3}}{2^{6}}+\ldots .\right] Since \left(1+\frac{3}{2^{2}}+\frac{3^{2}}{2^{4}}+_{--}-\right) is a G.P. therfore by sum of infinite G.P, we have =\frac{1}{2}\left[\frac{1}{1-\frac{3}{2^{2}}}\right]=\frac{1}{2}[11−34]=2\left[\frac{1}{1-\frac{3}{4}}\right]=2\therefore \quadGivenexpressionGiven expression=−1[\b because 1+ω+ω2=0]\begin{array}{l} =-1 \\ {\left[\b\text{ because }1+\omega+\omega^{2}=0\right]} \end{array}$
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