A) \[{{2}^{\frac{31}{32}}}\]
B) \[{{2}^{\frac{15}{16}}}\]
C) 1
D) 0
Correct Answer: A
Solution :
Given that, \[\sqrt{2\sqrt{2\sqrt{2\sqrt{2\sqrt{2}}}}}\] By shortcut method, \[x\frac{{{2}^{n-1}}}{{{2}^{n}}}={{2}^{\frac{{{2}^{\,\,5-1}}}{{{2}^{5}}}}}\] [Here \[x=2\]] \[={{2}^{\frac{2\times 2\times 2\times 2\times 2-1}{2\times 2\times 2\times 2\times 2}}}={{2}^{\frac{31}{32}}}\]
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