Question

$2^{1/4}.2^{2/8}.2^{3/16}.2^{4/32}......\infty$ is equal to-

• A. 1
• B. 2
• C. 3/2
• D. 5/2

Answer 1

Answer: (B)

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 $2$