Question

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

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

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)
$S_{\infty }=\frac{a}{1-r}=2$
$\therefore {\omega }^1+{\omega }^2=-1$