Q84
ArithmeticNumber Series & Progressions
Find the value of $\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)\left(1-\dfrac{1}{5}\right)\cdots\left(1-\dfrac{1}{99}\right)\left(1-\dfrac{1}{100}\right)$
- a..01
- b..02
- c.$\dfrac{1}{2}$
- d..03
Answer: (B) .02
Each bracket has the form $1-\dfrac{1}{n}=\dfrac{n-1}{n}$, so the product is $\dfrac{2}{3}\times\dfrac{3}{4}\times\dfrac{4}{5}\times\cdots\times\dfrac{98}{99}\times\dfrac{99}{100}$. Every numerator cancels with the denominator just before it (a telescoping product), leaving only the first numerator and the last denominator: $\dfrac{2}{100}=0.02$. Note that the product starts at $\dfrac{1}{3}$, not $\dfrac{1}{2}$; if it had started at $\dfrac{1}{2}$ the answer would have been $\dfrac{1}{100}=0.01$, which is the trap in option (A).