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  4. 2(1/4), 4(1/8), 8(1/16), 16(1/32)............ is equal to

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Q. $2^{\frac{1}{4}}, 4^{\frac{1}{8}}, 8^{\frac{1}{16}}, 16^{\frac{1}{32}}............ $ is equal to

Sequences and Series

Solution:

$2^{\frac{1}{4}}.4^{\frac{1}{8}} .8^{\frac{1}{16}} .16^{\frac{1}{32}}.....$
$ = 2^{\frac{1}{4}}.2^{\frac{2}{8}}.2^{\frac{3}{16}}.2^{\frac{4}{32}} ......$
$= 2^{\frac{1}{4}+\frac{2}{8}+\frac{3}{16}+\frac{4}{32}} + ..... = 2^{s}$
where $S = \frac{1}{4}+\frac{2}{8}+\frac{3}{16}+\frac{4}{32}+.....$
$\therefore \frac{1}{2}S = \frac{1}{8}+\frac{2}{16}+\frac{3}{32} +..... $
subtracting, we get
$ \frac{S}{2} = \frac{1}{4} +\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+.....\infty $
$ = \frac{\frac{1}{4}}{1-\frac{1}{2}} = \frac{\frac{1}{4}}{\frac{1}{2}} = \frac{1}{2} $
$ \Rightarrow S = 1$.
$ \therefore $ reqd sum $2^{1} = 2$.