Question

$2^{\frac{1}{4}}·4^{\frac{1}{8}}·8^{\frac{1}{16}}·{16}^{\frac{1}{32}}·.........$is equal to

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

Answer 1

The correct option is B

$2$

Explanation for correct option:

Step 1: Add powers of $2$ in given series

Let the given series is

$$\begin{gathered} \mathrm{S}=2^{\frac{1}{4}}·4^{\frac{1}{8}}·8^{\frac{1}{16}}·{16}^{\frac{1}{32}}·......... \\ \Rightarrow \mathrm{S}=2^{\frac{1}{4}}·{\left(2\right)}^{2\times \frac{1}{8}}·{\left(2\right)}^{3\times \frac{1}{16}}·{\left(2\right)}^{4\times \frac{1}{32}}·......... \\ \Rightarrow \mathrm{S}=2^{\frac{1}{4}+\frac{2}{8}+\frac{3}{16}+\frac{4}{32}+...........} \end{gathered}$$

Let $k=\frac{1}{4}+\frac{2}{8}+\frac{3}{16}+\frac{4}{32}+.............\left(1\right)$

Divided by $2$ both the side.

$\Rightarrow \frac{k}{2}=\frac{1}{8}+\frac{2}{16}+\frac{3}{32}+\frac{4}{64}+..............\left(ii\right)$

Subtract from equation $\left(i\right)$ to equation $\left(ii\right)$

$\frac{k}{2}=\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+......$

Step 2: Apply formula for sum of infinite G.P.

Sum of infinite G.P. is $S_{\infty }=\frac{a}{1-r}$

Here in series $a=\frac{1}{4}\&r=\frac{1}{2}$

$$\begin{gathered} \Rightarrow \frac{k}{2}=\frac{{\displaystyle \frac{1}{4}}}{1-{\displaystyle \frac{1}{2}}} \\ \Rightarrow \frac{k}{2}=\frac{1}{2} \\ \Rightarrow k=1 \end{gathered}$$

Now, $S=2^1=2$

Therefore, correct answer is option $\left(B\right)$