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