A) 1
B) 2
C) 3/2
D) 4
Correct Answer: B
Solution :
\[{{2}^{1/4}}{{.4}^{1/8}}{{.8}^{1/6}}.....\infty \] \[={{2}^{\frac{1}{4}+\frac{2}{8}+\frac{3}{16}+.....\infty }}\] \[=\frac{1}{{{2}^{2}}}\left\{ \frac{1}{1-1/2}+\frac{1-1/2}{{{\left( 1-\frac{1}{2} \right)}^{2}}} \right\}\] \[={{2}^{1/{{2}^{2}}|2+2|}}={{2}^{1}}=2\]You need to login to perform this action.
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