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1. The value of $\sqrt{2 \sqrt{2 \sqrt{2 \sqrt{2 \sqrt{2}}}}} = ?$

A. $2^{\frac{29}{31}}$

B. $2^{\frac{31}{32}}$

C. $2^{\frac{9}{2}}$

D. $2^{\frac{11}{2}}$

Solution:

\begin{aligned} \sqrt{2 \sqrt{2 \sqrt{2 \sqrt{2 \sqrt{2}}}}} \\ = \sqrt{2 \sqrt{2 \sqrt{2 \sqrt{{2.2}^{\frac{1}{2}}}}}} \\ = \sqrt{2 \sqrt{2 \sqrt{2 \sqrt{2^{(1 + \frac{1}{2})}}}}} \\ = \sqrt{2 \sqrt{2 \sqrt{2 \sqrt{2^{\frac{3}{2}}}}}} \\ = \sqrt{2 \sqrt{2 \sqrt{2. {(2^{\frac{3}{2}})}^{\frac{1}{2}}}}} \\ = \sqrt{2 \sqrt{2 \sqrt{{2.2}^{(\frac{3}{2} \times \frac{1}{2})}}}} \\ = \sqrt{2 \sqrt{2 \sqrt{{2.2}^{\frac{3}{4}}}}} \\ = \sqrt{2 \sqrt{2 \sqrt{2^{(1 + \frac{3}{4})}}}} \\ = \sqrt{2 \sqrt{2 \sqrt{2^{\frac{7}{4}}}}} \\ = \sqrt{2 \sqrt{2 {(2^{\frac{7}{4}})}^{\frac{1}{2}}}} \\ = \sqrt{2 \sqrt{{2.2}^{(\frac{7}{4} \times \frac{1}{2})}}} \\ = \sqrt{2 \sqrt{{2.2}^{\frac{7}{8}}}} \\ = \sqrt{2 \sqrt{2^{(1 + \frac{7}{8})}}} \\ = \sqrt{2 \sqrt{2^{\frac{15}{8}}}} \\ = \sqrt{2 {(2^{\frac{15}{8}})}^{\frac{1}{2}}} \\ = \sqrt{{2.2}^{\frac{15}{16}}} \\ = \sqrt{2^{(1 + \frac{15}{16})}} \\ = \sqrt{2^{\frac{31}{16}}} \\ = {(2^{\frac{31}{16}})}^{\frac{1}{2}} \\ = 2^{(\frac{31}{16} \times \frac{1}{2})} \\ = 2^{\frac{31}{32}} \end{aligned}

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