If $\omega$ is a complex cube root of unity, then $\omega^{\left(\frac{1}{3}+\frac{2}{9}+\frac{4}{27}+\ldots…
If $\omega$ is a complex cube root of unity, then $\omega^{\left(\frac{1}{3}+\frac{2}{9}+\frac{4}{27}+\ldots \infty\right)}+\omega^{\left(\frac{1}{2}+\frac{3}{8}+\frac{9}{32}+\ldots \infty\right)}$ is equal to
$1$
$-1$
$\omega$
$i$
Solution
We have, $\omega^{\left(\frac{1}{3}+\frac{2}{9}+\frac{4}{27}+\ldots . \infty\right)}+\omega^{\left(\frac{1}{2}+\frac{3}{8}+\frac{9}{32}+\ldots . \infty\right)}=$ ?
Here, $\frac{1}{3}+\frac{2}{9}+\frac{4}{27}+\ldots \infty$
Follows infinite G.P. series so its sum,
$\begin{aligned}
& S_{\infty}=\frac{a}{1-r}=\frac{\frac{1}{3}}{1-\frac{2}{3}} \\
& =\frac{1}{3} \times \frac{3}{1}=1 \quad \therefore \omega^{\left(\frac{1}{3}+\frac{2}{9}+\frac{4}{27}+\ldots \infty\right)}=\omega
\end{aligned}$
Now, $\frac{1}{2}+\frac{3}{8}+\frac{9}{32}+\ldots . .+\infty$
also follows infinite G.P. series so its sum
$\begin{aligned} & \mathrm{S}_{\propto}=\frac{a}{1-r}=\frac{\frac{1}{2}}{1-\frac{3}{4}}=\frac{1}{2} \times \frac{4}{1}=2 \\ & \therefore \quad \omega^{\left(\frac{1}{2}+\frac{3}{8}+\ldots+\infty\right)}=\omega^2 \\ & \text { Hence, } \quad \omega^{\left(\frac{1}{3}+\frac{2}{9}+\ldots+\infty\right)}+\omega^{\left(\frac{1}{2}+\frac{3}{8}+\ldots+\infty\right)}=\omega+\omega^2=-1 \\ & {\left[\because \omega^2+\omega+1=0 \Rightarrow \omega^2+\omega=-1\right]} \\ & \end{aligned}$