Question

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

• A. 1
• B. -1
• C. $\omega$
• D. $\iota$

Answer 1

The correct option is B -1

$\frac{1}{3}+\frac{2}{9}+\frac{4}{27}+.....+\infty$(infinite G.P.)
$S_{\infty }=\frac{a}{1-r}=1$
$\frac{1}{2}+\frac{3}{8}+\frac{9}{32}+.....+\infty$(infinite G.P.)
$S_{\infty }=\frac{a}{1-r}=2$
$\therefore {\omega }^1+{\omega }^2=-1$