If omega is a complex cube root of unity,then | vedclass

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(B) The first exponent is an infinite geometric series with $a = \frac{1}{3}$ and $r = \frac{2}{3}$.Its sum is $S_{\infty} = \frac{a}{1-r} = \frac{1/3

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$-1$

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(B) The first exponent is an infinite geometric series with $a = \frac{1}{3}$ and $r = \frac{2}{3}$.Its sum is $S_{\infty} = \frac{a}{1-r} = \frac{1/3}{1-2/3} = \frac{1/3}{1/3} = 1$.Thus,$\omega^{\left(\frac{1}{3}+\frac{2}{9}+\ldots\right)} = \omega^1 = \omega$.The second exponent is an infinite geometric series with $a = \frac{1}{2}$ and $r = \frac{3}{4}$.Its sum is $S_{\infty} = \frac{a}{1-r} = \frac{1/2}{1-3/4} = \frac{1/2}{1/4} = 2$.Thus,$\omega^{\left(\frac{1}{2}+\frac{3}{8}+\ldots\right)} = \omega^2$.Adding these results,we get $\omega + \omega^2$.Since $1 + \omega + \omega^2 = 0$,it follows that $\omega + \omega^2 = -1$.

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

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