Question

The value of ${(1-\frac{1}{3})}^2{(1-\frac{1}{4})}^2{(1-\frac{1}{5})}^2.....{(1-\frac{1}{n})}^2$ is equal to

• A. ${\left(\frac{1}{n}\right)}^2$
• B. ${\left(\frac{2}{n}\right)}^2$
• C. ${\left(\frac{3}{n}\right)}^2$
• D. ${\left(\frac{4}{n}\right)}^2$

Answer 1

The correct option is B ${\left(\frac{2}{n}\right)}^2$

${(1-\frac{1}{3})}^2{(1-\frac{1}{4})}^2{(1-\frac{1}{5})}^2.....{(1-\frac{1}{n})}^2$
$\Rightarrow {\left(\frac{2}{3}\right)}^2{\left(\frac{3}{4}\right)}^2{\left(\frac{4}{5}\right)}^2........{\left(\frac{n-1}{n}\right)}^2$
$\Rightarrow \frac{4}{9}\times \frac{9}{16}\times \frac{16}{25}.......\frac{(n-1{)}^2}{(n{)}^2}$
$\Rightarrow {\left(\frac{2}{n}\right)}^2$