Question

The product of the numbers $2^{1/4} \cdot4^{1/8} \cdot8^{1/16} \cdot16^{1/32}$ ... is

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

Answer 1

Answer: (D)

Product $= 2^{1/4} 2^{2/8} 2^{3/16} 2^{4/32} 2^{5/64} ...$
$ = 2^{\frac{1}{4} +\frac{2}{8} +\frac{3}{16} +\frac{4}{32} +\frac{5}{64} + ....\infty} ....\left(*\right)$
$P =2^{R}$, where $R = \frac{1}{4} +\frac{2}{8} +\frac{3}{16} +\frac{4}{32}+ ....\left(i\right)$
$\therefore \frac{R}{2} =\frac{1}{8} +\frac{2}{16} +\frac{3}{32} + ...\left(ii\right)$
$\therefore $ (i) -(ii) gives
$R - \frac{R}{2} = \frac{1}{4} +\frac{1}{8} +\frac{1}{16} +\frac{1}{32} + ...= \frac{1/4}{1 -1/2}$
or$\frac{R}{2} = \frac{1}{4} \times\frac{2}{1}$ or $R =1$
$\therefore $ Product $= 2^{R} = 2^{1} = 2$