Question

$2^{\frac{1}{4}}, 4^{\frac{1}{8}}, 8^{\frac{1}{16}}, 16^{\frac{1}{32}}............$ is equal to

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

Answer 1

Answer: (B)

2

Answer 2

$2^{\frac{1}{4}}.4^{\frac{1}{8}} .8^{\frac{1}{16}} .16^{\frac{1}{32}}.....$ $= 2^{\frac{1}{4}}.2^{\frac{2}{8}}.2^{\frac{3}{16}}.2^{\frac{4}{32}} ......$ $= 2^{\frac{1}{4}+\frac{2}{8}+\frac{3}{16}+\frac{4}{32}} + ..... = 2^{s}$ where $S = \frac{1}{4}+\frac{2}{8}+\frac{3}{16}+\frac{4}{32}+.....$ $\therefore \frac{1}{2}S = \frac{1}{8}+\frac{2}{16}+\frac{3}{32} +.....$ subtracting, we get $\frac{S}{2} = \frac{1}{4} +\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+.....\infty$ $= \frac{\frac{1}{4}}{1-\frac{1}{2}} = \frac{\frac{1}{4}}{\frac{1}{2}} = \frac{1}{2}$ $\Rightarrow S = 1$. $\therefore$ reqd sum $2^{1} = 2$.