Question

The product $2^{\frac{1}{4}}\times 4^{\frac{1}{16}}\times 8^{\frac{1}{48}}\times {16}^{\frac{1}{128}}...$ to $\infty$ is equal to :

• A. $2^{\frac{1}{4}}$
• B. $2$
• C. $2^{\frac{1}{2}}$
• D. $1$

Answer 1

The correct option is C $2^{\frac{1}{2}}$

$2^{\frac{1}{4}}\times 4^{\frac{1}{16}}\times 8^{\frac{1}{48}}...\infty =2^{\frac{1}{4}}\times 2^{\frac{2}{16}}\times 2^{\frac{4}{48}}\times ...\infty$
$\Rightarrow 2^{\frac{1}{4}}\times 2^{\frac{1}{8}}\times 2^{\frac{1}{16}}...\infty =2^{\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+...\infty }$
$\Rightarrow 2^{\left(\frac{\frac{1}{4}}{1-\frac{1}{2}}\right)}=2^{\frac{1}{2}}$