Question

The value of $2^{\frac{1}{4}}{.4}^{\frac{1}{8}}{.8}^{\frac{1}{16}}{.16}^{\frac{1}{32}}....\infty$ equals

• A. $1$
• B. $2$
• C. $\frac{3}{2}$
• D. $4$

Answer 1

The correct option is B $2$

$2^{\frac{1}{4}}.4^{\frac{1}{8}}.8^{\frac{1}{16}}.{16}^{\frac{1}{32}}................\infty$

$=2^{\frac{1}{4}}.2^{\frac{2}{8}}.2^{\frac{3}{16}}.2^{\frac{4}{32}}............\infty$

$=2^{(\frac{1}{4}+\frac{2}{8}+\frac{3}{16}+\frac{4}{32}+..........\infty )}=2^p$

So let us calculate $p$.

$p=\frac{1}{4}+\frac{2}{8}+\frac{3}{16}+\frac{4}{32}+..........\infty$

$\Rightarrow \frac{1}{2}p=\frac{1}{8}+\frac{2}{16}+\frac{3}{32}+\frac{4}{64}+..............\infty$

On subtracting the two equations we get

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

$\Rightarrow \frac{1}{2}p=\frac{\frac{1}{4}}{1-\frac{1}{2}}=\frac{2}{4}=\frac{1}{2}\Rightarrow p=1$

Hence the value=$2^p=2$