A) 1
B) 2
C) \[\frac{3}{2}\]
D) \[\frac{5}{2}\]
Correct Answer: B
Solution :
\[{{2}^{1/4}}{{.4}^{1/8}}{{.8}^{1/16}}{{.16}^{1/32}}.....\infty \] \[={{2}^{1/4+2/8+3/16+......}}={{2}^{S}}\], where \[S\] is given by \[S=\frac{1}{4}+2\frac{1}{8}+3\frac{1}{16}+4\frac{1}{32}+...........\infty \] ......(i) \[\Rightarrow \]\[\frac{1}{2}S=\text{ }\frac{1}{8}+\frac{2}{16}+\frac{3}{32}+\frac{4}{64}+..........\infty \] ......(ii) Subtracting (ii) from (i), we get \[S=1\]. Hence required product\[={{2}^{1}}=2\].
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