Question

Tính \(A=\left(1-\frac{1}{1+2}\right).\left(1-\frac{1}{1+2+3}\right).......\left(1-\frac{1}{1+2+3+...+2006}\right)\)

Answer 1

\(A=\left(1-\frac{1}{1+2}\right).\left(1-\frac{1}{1+2+3}\right)......\left(1-\frac{1}{1+2+3+...+2006}\right)\)

\(A=\left(1-\frac{1}{3}\right)\left(1-\frac{1}{6}\right)\left(1-\frac{1}{10}\right)....\left(1-\frac{1}{2013021}\right)\)

\(A=\frac{2}{3}.\frac{5}{6}.\frac{9}{10}....\frac{2013020}{2013021}\)

\(A=\frac{4}{6}.\frac{10}{12}.\frac{18}{20}......\frac{4026040}{4026042}\)

\(A=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}......\frac{2005.2008}{2006.2007}\)

\(A=\frac{1.2.3.....2005}{2.3.4....2006}.\frac{4.5.6....2008}{3.4.5...2007}\)

\(A=\frac{1}{2006}.\frac{2008}{3}=\frac{1004}{3009}\)