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The value of $2^{1/4}\: \: .4^{1/8}\: \: .8^{1/6}\: \: .....\infty$ is

Option 1)
$1$

Option 2)
$2$

Option 3)
$3/2$

Option 4)
$4$

Answers (1)

best_answer

As we learnt in

Sum of infinite terms of a GP -

$a+ar+ar^{2}+- - - - -= \frac{a}{1-r}\\here \left | r \right |<1$

- wherein

$a\rightarrow$ first term

$r\rightarrow$ common ratio

and

Sum of infinite terms of an AGP -

$S_{\infty }= \frac{a}{1-r}+\frac{dr}{\left ( 1-r \right )^{2}}\\here \: \left | r \right |< 1$

- wherein

$a\rightarrow$ first term

$d\rightarrow$ common diff. of a AP

$r\rightarrow$ common ratio of a GP

$2\tfrac{1}{4}\cdot 4\tfrac{1}{8}\cdot 8\tfrac{1}{16}\cdot \cdot \infty$

$= 2\tfrac{1}{4}\cdot 2\tfrac{2}{8}\cdot 2\tfrac{3}{16}\cdot \cdot \cdot \infty$

$= 2^{\tfrac{1}{4}+\frac{2}{8}+\frac{3}{16}+\cdot \cdot \infty}$

Now let $S= \frac{1}{4}+\frac{2}{8}+\frac{3}{16}+\cdot \cdot \infty -(i)$

$\frac{S}{2}= \frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\cdot \cdot$ (ii)

Subtracting (ii) from (i)

We get

$\therefore\ \; \frac{s}{2}=\frac{1}{4}+\frac{1}{8}+\frac{1}{16}..........$

$= \frac{\frac{1}{4}}{1-\frac{1}{2}}= \frac{1}{4}\times \frac{2}{1}= \frac{1}{2}$

$\therefore S=1$

From (i)

$\therefore 2^{1}=2$

Option 1)

$1$

This option is incorrect.

Option 2)

$2$

This option is correct.

Option 3)

$3/2$

This option is incorrect.

Option 4)

$4$

This option is incorrect.

Posted by

divya.saini

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