\(\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot...\cdot\left(1-\frac{1}{100}\right)\)
\(=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot....\cdot\frac{99}{100}\)
\(=\frac{1\cdot2\cdot3\cdot...\cdot99}{2\cdot3\cdot4\cdot...\cdot100}\)
\(=\frac{1}{100}\)
\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{100}\right)=\frac{2-1}{2}.\frac{3-1}{3}.\frac{4-1}{4}...\frac{100-1}{100}=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{99}{100}=\frac{1}{100}\)
\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{100}\right)\)
\(\Rightarrow\frac{1}{2}.\frac{2}{3}.\frac{3}{4}....\frac{99}{100}\)
\(\Rightarrow\frac{1.2.3.4.5......99}{2.3.4.....100}\)
Áp dụng tính chất loại bỏ dần ta được kết quả.
\(=\frac{1}{100}\)
\(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{100}\right)\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{99}{100}\)
\(=\frac{1.2.3...99}{2.3.4...100}\)
\(=\frac{1}{100}\)