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)

$2^{\frac{1}{4}}.4^{\frac{1}{8}}.8^{\frac{1}{16}}.16^{\frac{1}{32}}.....$
$=2^{\frac{1}{4}}.2^{\frac{2}{8}}.2^{\frac{3}{16}}.2^{\frac{4}{32}}......$
$=2^{\frac{1}{4}+\frac{2}{8}+\frac{3}{16}+\frac{4}{32}}+.....=2^s$
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 $2^1=2$.