Question

The value of $\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)\left(1-\frac{1}{5}\right)\left(1-\frac{1}{6}\right).....\left(1-\frac{1}{n}\right)$ is _________.

• A. $\frac{1}{n}$
• B. $\frac{2}{n}$
• C. $\frac{2(n-1)}{4}$
• D. $\frac{2}{n(n+1)}$

Answer 1

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

We need to find value of $(1-\frac{1}{3})(1-\frac{1}{4})(1-\frac{1}{5})(1-\frac{1}{6}).....(1-\frac{1}{n})$

Solving each bracket, we get

$\frac{2}{3}\times \frac{3}{4}\times \frac{4}{5}.......\frac{n-3}{n-2}\times \frac{n-2}{n-1}\times \frac{n-1}{n}$

Starting from $1^{st}$ term each denominator is cancelled out by next numerator.

Finally, we get $\frac{2}{n}$