Question

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

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

Answer 1

Answer: (B)

$-1$

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