Question

$2^{\frac{1}{4}}{4}^{\frac{1}{8}}{8}^{\frac{1}{16}}{16}^{\frac{1}{32}}$ ... upto $\infty$, equals to :

• A. $1$
• B. $2$
• C. $\frac{3}{2}$
• D. $\frac{5}{2}$

Answer 1

The correct option is B $2$

Given:

$S=2^{\frac{1}{4}}{4}^{\frac{1}{8}}{8}^{\frac{1}{16}}\cdots \infty$

$\Rightarrow S=2^{\frac{1}{4}}{2}^{\frac{2}{8}}{2}^{\frac{3}{16}}\cdots \infty$

$\Rightarrow S=2^{\frac{1}{4}+\frac{2}{8}+\frac{3}{16}\cdots \infty }$

Divide both sides by $\sqrt{S}$, we get

$S^{1-\frac{1}{2}}=2^{(\frac{1}{4}+\frac{2}{8}+\frac{3}{16}+\cdots \infty )-(\frac{1}{8}+\frac{2}{16}+\cdots \infty )}$

$\Rightarrow S^{\frac{1}{2}}=2^{\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\cdots \infty }$

$\Rightarrow S^{\frac{1}{2}}=2^{\frac{1/4}{1-1/2}}=2^{\frac{1}{2}}$

$\Rightarrow S=2$