Question

If $E=\frac{1}{4}.\frac{2}{6}.\frac{3}{8}.\frac{4}{10}...\frac{30}{62}\frac{31}{64}=8^x$ then value of x is:

Answer 1

$IfE=\frac{1}{4}\times \frac{2}{6}\times \frac{3}{8}\times \frac{4}{10}.....\times \frac{30}{62}\times \frac{31}{64}=\frac{1}{4}\times \frac{1}{3}\times \frac{3}{8}\times \frac{2}{5}\times \frac{5}{12}\times \frac{3}{7}....\times \frac{15}{31}\times \frac{31}{64}=\frac{1}{4}\times \frac{1}{\text{/}3}\times \frac{\text{/}3}{8}\times \frac{2}{\text{/}5}\times \frac{\text{/}5}{12}\times \frac{3}{\text{/}7}-1....\times \frac{15}{\text{/}31}\times \frac{\text{/}31}{64}=\frac{1}{4}\times 1\times \frac{1}{8}\times \frac{2}{1}\times \frac{1}{12}\times \frac{3}{1}....\times 15\times \frac{1}{64}=\frac{1}{4}\times 1\times \frac{1}{\text{/}{9}_4}\times \frac{\text{/}2}{1}\times \frac{1}{\text{/}{12}_4}\times \frac{\text{/}3}{1}\times .....\times \text{/}15\times \frac{1}{64}$

$=(\frac{1}{4}{)}^{15}\times \frac{1}{64}=(\frac{1}{2^2}{)}^{15}\times \frac{1}{2^6}$