Question

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

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

Answer 1

The correct option is C

$2^{\frac{1}{2}}$

The explanation for the correct option.

Step $1$: Identify the sequence.

The given expression: $2^{\frac{1}{4}}\times 4^{\frac{1}{16}}\times 8^{\frac{1}{48}}\times {16}^{\frac{1}{128}}......\infty$

$$\begin{gathered} \Rightarrow 2^{\frac{1}{4}}\times 4^{\frac{1}{16}}\times 8^{\frac{1}{48}}\times {16}^{\frac{1}{128}}......\infty =2^{\frac{1}{4}}\times 2^{2\times \frac{1}{16}}\times 2^{3\times \frac{1}{48}}\times 2^{4\times \frac{1}{128}}......\infty \\ \Rightarrow 2^{\frac{1}{4}}\times 4^{\frac{1}{16}}\times 8^{\frac{1}{48}}\times {16}^{\frac{1}{128}}......\infty =2^{\frac{1}{4}}\times 2^{\frac{1}{8}}\times 2^{\frac{1}{16}}\times 2^{\frac{1}{32}}......\infty \\ \Rightarrow 2^{\frac{1}{4}}\times 4^{\frac{1}{16}}\times 8^{\frac{1}{48}}\times {16}^{\frac{1}{128}}......\infty =2^{\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+......\infty } \\ \Rightarrow 2^{\frac{1}{4}}\times 4^{\frac{1}{16}}\times 8^{\frac{1}{48}}\times {16}^{\frac{1}{128}}......\infty =2^{\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}+......\infty } \end{gathered}$$

The infinite progression is $\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}+......\infty$

The first term of the sequence, $a=\frac{1}{4}$.

The common ratio of the sequence, $r=\frac{a_2}{a_1}=\frac{{\displaystyle \frac{1}{2^3}}}{{\displaystyle \frac{1}{2^2}}}=\frac{2^2}{2^3}=\frac{1}{2}$.

Step $2$: Find sum of the infinite progression.

Since, sum of the infinite G.P. is given by, $\frac{a}{1-r}$

Thus,

$\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}+......\infty =\frac{a}{1-r}$.

$$\begin{gathered} \Rightarrow \frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}+......\infty =\frac{{\displaystyle \frac{1}{4}}}{1-{\displaystyle \frac{1}{2}}} \\ \Rightarrow \frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}+......\infty =\frac{{\displaystyle \frac{1}{4}}}{{\displaystyle \frac{1}{2}}} \\ \Rightarrow \frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}+......\infty =\frac{2}{4} \\ \Rightarrow \frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}+......\infty =\frac{1}{2} \end{gathered}$$

Step $3$: Find the required value.

Hence, $2^{\frac{1}{4}}\times 4^{\frac{1}{16}}\times 8^{\frac{1}{48}}\times {16}^{\frac{1}{128}}......\infty =2^{\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}+......\infty }$

$\Rightarrow 2^{\frac{1}{4}}\times 4^{\frac{1}{16}}\times 8^{\frac{1}{48}}\times {16}^{\frac{1}{128}}......\infty =2^{\frac{1}{2}}$

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

Hence option (C) is the correct option.