What is the value of \(\left(1-\frac{1}{4}\right)\left(1-\frac{1}{9}\right)\left(1-\frac{1}{16}\right)\cdots\left(1-\frac{1}{100}\right)\)?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\frac{11}{20}\)
The brackets are \(1-\frac{1}{k^{2}}\) for \(k=2,3,\ldots,10\). Each one splits as a difference of squares, \(\displaystyle1-\frac{1}{k^{2}}=\frac{k-1}{k}\times\frac{k+1}{k}\), and in the long product almost every factor cancels with the next one.
First factors: \(\displaystyle\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\times\cdots\times\frac{9}{10}=\frac{1}{10}\)
Second factors: \(\displaystyle\frac{3}{2}\times\frac{4}{3}\times\frac{5}{4}\times\cdots\times\frac{11}{10}=\frac{11}{2}\)
Product \(\displaystyle=\frac{1}{10}\times\frac{11}{2}=\frac{11}{20}\)