Question

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

Answer 1

$2^{\frac{1}{4}}\times 2^{\frac{2}{8}}\times 2^{\frac{3}{16}}\times 2^{\frac{4}{32}}\times 2^{\frac{5}{64}}......$

$P=(2{)}^{(\frac{1}{4}+\frac{2}{8}+\frac{3}{16}+\frac{4}{32}+\frac{5}{64}+....)}$

$P=(2{)}^S$ ……………… $\left(1\right)$

Where $S=\frac{1}{4}+\frac{2}{8}+\frac{3}{16}+\frac{4}{32}+\frac{5}{64}+....\infty$ ..... $\left(2\right)$

$S\times \left(\frac{1}{2}\right)=\frac{1}{8}+\frac{2}{16}+\frac{3}{32}+\frac{4}{64}+....+\infty$ ……. $\left(3\right)$

On subtracting equation $\left(3\right)$ from $\left(2\right)$, we get

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

$\frac{S}{2}=\frac{a}{1-r}=\frac{\frac{1}{4}}{1–\frac{1}{2}}$.

$\frac{S}{2}=\frac{1}{4}\times \frac{2}{1}.$

$\frac{S}{2}=\frac{1}{2}$

$S=1$

Put $S=1$ in eq.$\left(1\right)$.

$P=(2{)}^S.$

$P=(2{)}^1$

$P=2$