Question : $(1-\frac{1}{5})(1-\frac{1}{6})(1-\frac{1}{7}).......(1-\frac{1}{100})$ is equal to:
Option 1: $0$
Option 2: $\frac{1}{25}$
Option 3: $\frac{1}{100}$
Option 4: $\frac{1}{50}$
Correct Answer: $\frac{1}{25}$
Solution : $(1-\frac{1}{5})(1-\frac{1}{6})(1-\frac{1}{7}).......(1-\frac{1}{100})$ = $(\frac{4}{5})(\frac{5}{6})(\frac{6}{7}).......(\frac{99}{100})$ Notice that the denominator of the first term cancels the numerator of the second term. So, $(\frac{4}{5})(\frac{5}{6})(\frac{6}{7}).......(\frac{99}{100})$ = $\frac{4}{100}$ = $\frac{1}{25}$ Hence, the correct answer is $\frac{1}{25}$.
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