A) 1
B) \[-1\]
C) 0
D) None of these
Correct Answer: B
Solution :
\[{{x}_{n}}=\left( \cos \frac{\pi }{{{2}^{n}}}+i\sin \frac{\pi }{{{2}^{n}}} \right)={{e}^{i\pi /{{2}^{n}}}}\] Put \[n=1,2,3,....\] \[{{x}_{1}}={{e}^{i\pi /2}},{{x}_{2}}={{e}^{i\pi /{{2}^{2}}}},{{x}_{3}}={{e}^{i\pi /{{2}^{3}}}},...........\] Now, \[{{x}_{1}}.{{x}_{2}}.{{x}_{3}}.....={{e}^{i\pi /{{2}^{2}}}}.{{e}^{i\pi /{{2}^{2}}}}.{{e}^{i\pi /{{2}^{3}}}},.....\] \[={{e}^{i\pi \left( \frac{1}{2}+\frac{1}{{{2}^{2}}}+\frac{1}{{{2}^{3}}}+....\infty \right)}}\] \[={{e}^{i\pi \left( \frac{1/2}{1-1/2} \right)}}\] \[={{e}^{i\pi (1)}}\] \[=\cos \pi +i\sin \theta =-1\]
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