Question

The value of $2^{1/4}.4^{1/8}. 8^{1/6}....\infty$ is

• A. 1
• B. 2
• C. $\frac{3}{2}$
• D. 4

Answer 1

Answer: (B)

2

Answer 2

$2^{1/4}. 4^{1/8}. 8^{1/10}........\infty = 2^{1/4}.\left(2^{2}\right)^{1/8}.\left(2^{3}\right)^{1/16} ........\infty$ $= 2^{1/4} .2^{2/8}.2^{3/16} .......\infty = 2^{1/4+2/8+3/16+.....\infty}$ $= 2^{S}$ where $S= \frac{1}{4} +\frac{2}{8} +\frac{3}{16} + .....\infty$ $\therefore\frac{S}{2}= \frac{1}{8}+\frac{2}{16} +....\infty$ $\therefore S = \left(1-\frac{1}{2}\right)= \frac{1}{4}+\frac{1}{8}+\frac{1}{16}+ .....\infty= \frac{\frac{1}{4}}{1-\frac{1}{2}} = \frac{1}{2}$ $\Rightarrow \frac{S}{2} = \frac{1}{2}$ $\Rightarrow S=1$ $\therefore$ given expression $= 2^{1}= 2$