Q42 / 125 · BITSAT 2015 · Mathematics
If $\omega$ 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
- Correct answerA-1
- B1
- C$-\mathrm{i}$
- Di
Solution
Consider $\frac{1}{2}+\frac{3}{8}+\frac{9}{32}+\frac{27}{128}+\ldots .$
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
$$
=\frac{1}{2}\left[\frac{1}{1-\frac{3}{2^{2}}}\right]=\frac{1}{2}\left[\frac{1}{1-\frac{3}{4}}\right]=2
$$
$\therefore \quad$ Given expression
$$
\begin{array}{l}
=-1 \\
{\left[\because 1+\omega+\omega^{2}=0\right]}
\end{array}
$$
Hence, the correct answer is option: 'A'.