A) \[\frac{1}{n}\]
B) \[\frac{3}{n}\]
C) \[\frac{2}{n}\]
D) \[\frac{n-1}{3}\]
Correct Answer: C
Solution :
\[\left( 1-\frac{1}{3} \right)\left( 1-\frac{1}{4} \right)...\left( 1-\frac{1}{n} \right)\] \[=\left( \frac{3-1}{3} \right)\left( \frac{4-1}{4} \right)...\left( \frac{n-1}{n} \right)\] \[=\frac{2}{3}\times \frac{3}{4}\times ....\times \frac{n-1}{n}=\frac{2}{n}\]
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