2^{1/4} cdot 4^{1/8} cdot 8^{1/16} cdot 16^{1 | vedclass

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(B) Let the given expression be $P = 2^{1/4} \cdot 4^{1/8} \cdot 8^{1/16} \cdot 16^{1/32} \cdots$Expressing all terms with base $2$:$P = 2^{1/4} \cdot

Vedclass · Accepted Answer

$2$

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(B) Let the given expression be $P = 2^{1/4} \cdot 4^{1/8} \cdot 8^{1/16} \cdot 16^{1/32} \cdots$Expressing all terms with base $2$:$P = 2^{1/4} \cdot (2^2)^{1/8} \cdot (2^3)^{1/16} \cdot (2^4)^{1/32} \cdots$$P = 2^{1/4 + 2/8 + 3/16 + 4/32 + \cdots} = 2^S$,where $S = \sum_{n=1}^{\infty} \frac{n}{2^{n+1}}$.$S = \frac{1}{4} + \frac{2}{8} + \frac{3}{16} + \frac{4}{32} + \cdots$ $(i)$$\frac{1}{2}S = \frac{1}{8} + \frac{2}{16} + \frac{3}{32} + \cdots$ $(ii)$Subtracting $(ii)$ from $(i)$:$S - \frac{1}{2}S = \frac{1}{4} + (\frac{2}{8} - \frac{1}{8}) + (\frac{3}{16} - \frac{2}{16}) + \cdots$$\frac{1}{2}S = \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \cdots$This is an infinite geometric series with $a = 1/4$ and $r = 1/2$.$\frac{1}{2}S = \frac{1/4}{1 - 1/2} = \frac{1/4}{1/2} = \frac{1}{2}$.Thus,$S = 1$.Therefore,$P = 2^1 = 2$.

$2^{1/4} \cdot 4^{1/8} \cdot 8^{1/16} \cdot 16^{1/32} \cdots$ is equal to

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