A) \[1\]
B) \[2\]
C) \[3\]
D) \[4\]
Correct Answer: B
Solution :
\[{{2}^{1/4}}\cdot {{4}^{1/8}}\cdot {{8}^{1/16}}...\infty \] \[={{2}^{\frac{1}{4}+\frac{2}{8}+\frac{3}{16}+...\infty }}\] \[={{2}^{\frac{1}{{{2}^{2}}}\left( 1+\frac{2}{2}+\frac{3}{{{2}^{2}}}+\frac{4}{{{2}^{3}}}+... \right)}}\] \[={{2}^{\frac{1}{{{2}^{2}}}\left( \frac{1}{1-\frac{1}{2}}-\frac{1\cdot \frac{1}{2}}{{{\left( 1-\frac{1}{2} \right)}^{2}}} \right)}}\] \[={{2}^{\frac{1}{{{2}^{2}}}|2+2|}}={{2}^{1}}=2\]
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