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)

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.

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