A) 1
B) \[\frac{1}{2}\]
C) 2
D) None of these
Correct Answer: A
Solution :
[a] \[2={{2}^{n}}\sin \alpha +\frac{1}{{{2}^{n}}\sin \alpha }\] \[{{2.2}^{n}}\sin \alpha ={{({{2}^{n}}\sin \alpha )}^{2}}+1\] \[{{({{2}^{n}}\sin \alpha -1)}^{2}}=0\] \[\sin \alpha =\frac{1}{{{2}^{n}}}\] for \[n=1,\] \[\sin {{\alpha }_{1}}=\frac{1}{2}\] for \[n=2,\] \[\sin {{\alpha }_{2}}=\frac{1}{4}\] for \[n=3,\] \[\sin {{\alpha }_{3}}=\frac{1}{8}\] \[S=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...upto\,\,\infty \] \[=\frac{\frac{1}{2}}{1-\frac{1}{2}}=\frac{\frac{1}{2}}{\frac{1}{2}}=1\]You need to login to perform this action.
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