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BITSAT 2015 Paper

Section: Mathematics

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Question : 82 of 150

Marks: +1, -0

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

\omega^{\frac{1}{2}+\frac{3}{8}+\frac{9}{32}+\frac{27}{128}+\cdots}

is

- 1
1
- i
i

Solution:

Consider

\frac{1}{2}+\frac{3}{8}+\frac{9}{32}+\frac{27}{128}+

...

Which can be written as

\frac{3^0}{2}+\frac{3^1}{8}+\frac{3^2}{32}+\frac{3^3}{128}+

+ ...

=

\frac{1}{2}

\left[ 1+\frac{3}{2^2}+\frac{3^2}{2^4}+\frac{3^3}{2^6}+\cdots \right]

Since

\left[ 1+\frac{3}{2^2}+\frac{3^2}{2^4}+\frac{3^3}{2^6}+\cdots \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}\left[ \frac{1}{1-\frac{3}{4}} \right]

= 2

∴ Given expression

= –1

[Since 1 + ω +

\omega^2

= 0]

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