Explanation

(b)
$2^{\frac{1}{4}}.4^{\frac{1}{8}}.8^{\frac{1}{16}}\ldots\infty = 2^{\frac{1}{4} + \frac{2}{8} + \frac{3}{16} + ...\infty}$
$= (2)^{\frac{1}{4}\left\{ 1 + \frac{2}{2} + \frac{3}{2^{2}} + \frac{4}{3^{2}} + ... \right\}}$
$= (2)^{\frac{1}{4}\left\{ \frac{1}{1 - \frac{1}{2}} + \frac{1.\frac{1}{2}}{\left( 1 - \frac{1}{2} \right)^{2}} \right\}}$
$= 2^{\frac{1}{4}(2 + 2)} = 2$