The product 2^{frac{1}{4}} cdot 4^{frac{1}{16 | vedclass

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Question

(A) The given expression is $P = 2^{\frac{1}{4}} \cdot 4^{\frac{1}{16}} \cdot 8^{\frac{1}{48}} \cdot 16^{\frac{1}{128}} \cdot \dots \infty$. Expressin

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The given expression is $P = 2^{\frac{1}{4}} \cdot 4^{\frac{1}{16}} \cdot 8^{\frac{1}{48}} \cdot 16^{\frac{1}{128}} \cdot \dots \infty$. Expressing all terms with base $2$: $P = 2^{\frac{1}{4}} \cdot (2^2)^{\frac{1}{16}} \cdot (2^3)^{\frac{1}{48}} \cdot (2^4)^{\frac{1}{128}} \cdot \dots$ $P = 2^{\frac{1}{4}} \cdot 2^{\frac{2}{16}} \cdot 2^{\frac{3}{48}} \cdot 2^{\frac{4}{128}} \cdot \dots$ $P = 2^{\frac{1}{4}} \cdot 2^{\frac{1}{8}} \cdot 2^{\frac{1}{16}} \cdot 2^{\frac{1}{32}} \cdot \dots$ Using the property $a^m \cdot a^n = a^{m+n}$,we get: $P = 2^{(\frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \frac{1}{32} + \dots)}$ The exponent is an infinite geometric series with first term $a = \frac{1}{4}$ and common ratio $r = \frac{1}{2}$. The sum of an infinite geometric series is $S = \frac{a}{1-r}$. $S = \frac{1/4}{1 - 1/2} = \frac{1/4}{1/2} = \frac{1}{2}$. Therefore,$P = 2^{\frac{1}{2}}$.

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

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