Question

The simplest value of $\frac{(1+\frac{1}{2})(1+\frac{1}{3})(1+\frac{1}{4}).....(1+\frac{1}{50})}{(1-\frac{1}{2})(1-\frac{1}{3})(1-\frac{1}{4})......(1-\frac{1}{50})}$ is

• A. $\frac{7}{32}$
• B. $\frac{4}{7}$
• C. $1275$
• D. $\frac{256}{343}$

Answer 1

The correct option is C $1275$

$\frac{(1+\frac{1}{2})(1+\frac{1}{3})(1+\frac{1}{4}).....(1+\frac{1}{50})}{(1-\frac{1}{2})(1-\frac{1}{3})(1-\frac{1}{4})......(1-\frac{1}{50})}$

$\Rightarrow \frac{\frac{3}{2}\times \frac{4}{3}\times \frac{5}{4}.......\frac{51}{50}}{\frac{1}{2}\times \frac{2}{3}\times \frac{3}{4}......\frac{49}{50}}$

$\Rightarrow \frac{\frac{51}{2}}{\frac{1}{50}}$

$\Rightarrow \frac{51}{2}\times \frac{50}{1}=1275$