Mathematics - Sequence & Series Question with Solution | TestHub

Consider $\frac{1}{2}+\frac{3}{8}+\frac{9}{32}+\frac{27}{128}+\dots .$ Which can be written as $$\frac{3^0}{2^1}+\frac{3^1}{2^3}+\frac{3^2}{2^5}+\frac{3^3}{2^7}+\dots .$$ $=\frac{1}{2}\left[1+\frac{3}{2^2}+\frac{3^2}{2^4}+\frac{3^3}{2^6}+\dots .\right]$ Since $\left(1+\frac{3}{2^2}+\frac{3^2}{2^4}{+}_{--}-\right)$ is a G.P. therfore by sum of infinite G.P, we have $$=\frac{1}{2}\left[\frac{1}{1-\frac{3}{2^2}}\right]=\frac{1}{2}\left[\frac{1}{1-\frac{3}{4}}\right]=2$$ $\therefore$ Given expression $$\begin{array}{l}=-1\\ \left[\because 1+\omega +{\omega }^2=0\right]\end{array}$$