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Let $z=a(\cos\large\frac{\pi}{5}$$+i\sin\large\frac{\pi}{5})$$\quad a\in R,|a| < 1$ then $S=z^{2015}+z^{2016}+z^{2017}+.......$ equals

$\begin{array}{1 1}(A)\;\large\frac{a^{2015}}{z-1}&(B)\;\large\frac{a^{2015}}{1-z}\\(C)\;\large\frac{z^{2015}}{1-a}&(D)\;\large\frac{z^{2015}}{a-1}\end{array} $

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We have $|z|=|a| < 1$,thus
$S=\large\frac{z^{2015}}{1-z}$
But $z^{2015}=a^{2015}[\cos(403)\pi+i\sin(403)\pi]$
$\Rightarrow -a^{2015}$
$\therefore S=\large\frac{a^{2015}}{z-1}$
Hence (A) is the correct answer.
answered Apr 10, 2014 by sreemathi.v
 

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