\(A=\left(1-\frac{1}{1+2}\right)\left(1-\frac{1}{1+2+3}\right)....\left(1-\frac{1}{1+2+...+100}\right)\)
\(A=\left(1-\frac{1}{3}\right)\left(1-\frac{1}{6}\right).....\left(1-\frac{1}{5050}\right)\)
\(A=\frac{2}{3}.\frac{5}{6}....\frac{5049}{5050}=\frac{4}{6}.\frac{10}{12}...\frac{10098}{10100}\)
\(A=\frac{1.4}{2.3}.\frac{2.5}{3.4}...\frac{99.102}{100.101}\)
\(A=\frac{1.2...98.99}{2.3...99.100}.\frac{4.5...102}{3.4...101}=\frac{1}{100}.\frac{102}{3}\)
Vậy \(A=\frac{102}{300}=\frac{17}{50}\)
\(A=\left(1-\frac{1}{1+2}\right).\left(1-\frac{1}{1+2+3}\right).....\left(1-\frac{1}{1+2+...+100}\right)\)
\(=\left(1-\frac{1}{3}\right)\left(1-\frac{1}{6}\right)...\left(1-\frac{1}{5050}\right)\)
\(=\frac{2}{3}.\frac{5}{6}.....\frac{5049}{5050}\)
\(=\frac{4}{6}.\frac{10}{12}.....\frac{10098}{10100}\)
\(=\frac{1.4}{2.3}.\frac{2.5}{3.4}.....\frac{99.102}{100.101}\)
\(=\frac{1.2.3..98.99}{2.3.4..99.100}.\frac{4.5.6...102}{3.4.5...101}\)
\(=\frac{1}{100}.\frac{102}{3}\)
\(=\frac{17}{50}\)
