Question

If ${}^{\prime }{\omega }^{\prime }$ is a 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 $-1$

KEY: 2
$\frac{1}{3}+\frac{2}{9}+\frac{4}{27}+...+\infty$ (infinte G.P.)

$S_{\infty }=\frac{a}{1-r}=1$

$\frac{1}{2}+\frac{3}{8}+\frac{9}{32}+...+\infty$ (infinte G.P.)
$S_{\infty }=\frac{a}{1-r}=2$

$\therefore {\omega }^1+{\omega }^2=-1$