Question

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

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

Answer 1

The correct option is B $2$

Converting all bases into bases of 2 we finally need to find sum of AGP
$S=\frac{1}{4}+\frac{2}{8}+\frac{3}{16}+\cdots$
Dividing by 2
$\frac{S}{2}=\frac{1}{8}+\frac{2}{16}+\frac{3}{32}+\cdots$
Subtracting
$\frac{S}{2}=\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\cdots$
Infinite GP
$\frac{S}{2}=\frac{\frac{1}{4}}{\left(1-\frac{1}{2}\right)}$
S=1 $\therefore$ the value of the expression is 2