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The product  2^{1/4}\cdot 4^{1/16}\cdot 8^{1/48}\cdot 16^{1/128}\cdot \cdots is equal to :   
Option: 1 2^{1/4}
Option: 2 2
Option: 3 2^{1/2}
Option: 4 1
 

Answers (2)

best_answer

Sum of an infinite GP   

 

If a is the first term and r is the common ratio of a G.P. Then,

\mathrm{S_{\infty}=\frac{a}{1-r}}

S_{\infty} is the sum to infinite terms of the G.P.


Now,

\\2^{1/4}\cdot4^{1/16}\cdot8^{1/48}\ldots=2^{\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\ldots}\\\Rightarrow 2^{\frac{\frac{1}{4}}{1-\frac{1}{2}}}=\sqrt 2

Posted by

avinash.dongre

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Option no 2

Posted by

Soham Rahate

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