A) 0
B) \[-1\]
C) 1
D) \[i\]
Correct Answer: C
Solution :
\[1+\sum\limits_{k=0}^{14}{\left\{ \cos \frac{2k+1}{15}\pi +i\sin \frac{(2k+1)}{15}\pi \right\}}\] \[=1+\sum\limits_{k=0}^{14}{{{e}^{i\frac{(2k+1)}{15}\pi }}}\] \[=1+(\alpha +{{\alpha }^{3}}+{{\alpha }^{5}}+....+{{\alpha }^{29}})\] where, \[\alpha ={{e}^{i\pi /15}}\] \[=1+\alpha \left[ \frac{1-{{({{\alpha }^{2}})}^{15}}}{1-{{\alpha }^{2}}} \right]\] \[=1+\alpha \left( \frac{1-{{\alpha }^{30}}}{1-{{\alpha }^{2}}} \right)\] \[=1+\alpha \left( \frac{1-1}{1-{{\alpha }^{2}}} \right)=1\] \[(\because \,{{\alpha }^{30}}={{e}^{i2\pi }}=1)\]
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