First slide

Geometric progression

Question

$The product 2^{\frac{1}{4}} \cdot 4^{\frac{1}{16}} \cdot 8^{\frac{1}{48}} \cdot 16^{\frac{1}{128}} \dots . . to \infty is equal to:$

Moderate

A

1
B

$2^{\frac{1}{2}}$
C

$2^{\frac{1}{4}}$
D

0

Solution

$\begin{array}{l} Sum of infinite geometric progression \frac{a}{1 - r} = \frac{\frac{1}{4}}{1 - \frac{1}{2}} = \frac{1}{2} \\ Therefore, 2^{\frac{1}{4}} \cdot 4^{\frac{1}{16}} \cdot 8^{\frac{1}{48}} \cdot 16^{\frac{1}{128}} \dots \dots to \infty = \sqrt{2} \end{array}$

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