Question

The value of $(1-\frac{1}{3})(1-\frac{1}{4})(1-\frac{1}{5})...(1-\frac{1}{n})$ is equal to

• A. $\frac{1}{n}$
• B. $\frac{2}{n}$
• C. $\frac{3}{n}$
• D. $\frac{4}{n}$

Answer 1

The correct option is B $\frac{2}{n}$

On simplification of first two term , we observed that
$\left(\frac{2}{3}\right)$ $\times \left(\frac{3}{4}\right)...$
So numerator is the multiplication from 2 to n-1 and denominator is the multiplication from 3 to n
So making it as factorial of first n and n-1 term we have to multiply 1 to numerator and 1 and 2 to denominator
$2\left(\frac{(n-1)!}{n!}\right)=\frac{2}{n}$