Question

If $\omega$ is a complex cube root of unity, then $\omega^{\left(\frac{1}{3}+\frac{2}{9}+\frac{4}{27}+\ldots \infty\right)}+\omega^{\left(\frac{1}{2}+\frac{3}{8}+\frac{9}{32}+\ldots \infty\right) \text { is equal to }}$

• A. 1
• B. -1
• C. $\omega$
• D. i

Answer 1

Answer: (B)

-1

Answer 2

Given that, $\omega^{\left(\frac{1}{3}+\frac{2}{9}+\frac{4}{27}+\ldots \infty\right)}+\omega^{\left(\frac{1}{2}+\frac{3}{8}+\frac{9}{32}+\ldots \infty\right)}$ Now, $\omega^{\left(\frac{1}{3}+\frac{2}{9}+\frac{4}{27}+\ldots \infty\right)}$ $\because \frac{1}{3}+\frac{2}{9}+\frac{4}{27}+\ldots \infty$ [infinite GP] $\therefore S_{\infty}=\frac{a}{1-r}=\frac{\frac{1}{3}}{1-\frac{2}{3}}$ $\left[\right.$ where, $\left.a=\frac{1}{3}, r=\frac{2}{9}+\frac{1}{3}=\frac{2}{9} \times \frac{3}{1}=\frac{2}{3}\right]$ $=\frac{1}{3} \times \frac{3}{1}=1$ and $\omega^{\left(\frac{1}{2}+\frac{3}{8}+\frac{9}{32}+\ldots \infty\right)}$ $\because \frac{1}{2}+\frac{3}{8}+\frac{9}{32}+\ldots .+\infty$ [infinite GP] $\therefore s_{\alpha}= \frac{a}{1-r} [\text { where, } a=\frac{1}{2}, r=\frac{3}{8} \times \frac{2}{1}=\frac{3}{4}]$ $=\frac{\frac{1}{2}}{1-\frac{3}{4}}=\frac{1}{2} \times \frac{4}{1}=2$ $\therefore \omega^{1}+\omega^{2}=-1$ $[\because \omega^{2}+\omega+1=0$ $\Rightarrow \omega^{2}+\omega=-1]$