Question

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

• A. 1
• B. 2
• C. 44622
• D. 44683

Answer 1

Answer: (B)

2

Answer 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 $2$