The Taylor series expansion of \frac{1}{z−2},|.z|. < 1 is

Correct Answer is : −12(1+z2+z24+z38...)

−\frac{1}{2} (1−\frac{z}{2} + \frac{z^{2}}{4}−\frac{z^{3}}{8}...)

−\frac{1}{2} (1 + \frac{z}{2} + \frac{z^{2}}{4} + \frac{z^{3}}{8}...)

−\frac{1}{2} (1−\frac{z}{2}−\frac{z^{2}}{4} + \frac{z^{3}}{8}...)

x−\frac{1}{2} (1 + \frac{z}{2}−\frac{z^{2}}{4}−\frac{z^{3}}{8}...)

AAI ATC Junior Executive 26 Dec 2015 Shift 2 Paper

Question : 111 of 120

Marks: +1, -0

The Taylor series expansion of

\;{1}/{z−2},|.z|. < 1

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