Question

$2^{1/4}{.4}^{1/8}{.8}^{1/16}{.16}^{1/32}$....... is equal to

Answer 1

Let $x=2^{1/4}{.4}^{1/8}{.8}^{1/16}{.16}^{1/32}......$

$x=2^{1/4}{.2}^{2/8}{.2}^{3/16}{.2}^{4/32}.....$

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

Let $y=\frac{1}{4}+\frac{2}{8}+\frac{3}{16}+\frac{4}{32}+.......$
$\frac{y}{2}=0+\frac{1}{8}+\frac{2}{16}+\frac{3}{32}+\frac{4}{64}+.......$

Subtracting both we get
$\frac{y}{2}=\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+.......$
$\frac{y}{2}=\frac{1}{4(1-\frac{1}{2})}=\frac{1}{2}$
$y=1$
Then, $x=2$