Đặt \(S=\left(1-\frac{1}{1+2}\right)\left(1-\frac{1}{1+2+3}\right)...\left(1-\frac{1}{1+2+3+...+2011}\right)\)
=\(\left(1-\frac{1}{2.\left(2+1\right):2}\right)\left(1-\frac{1}{3.\left(3+1\right):2}\right)...\left(1-\frac{1}{2011.\left(2011+1\right):2}\right)\)
=\(\left(1-\frac{2}{2.\left(2+1\right)}\right)\left(1-\frac{2}{3\left(3+1\right)}\right)...\left(1-\frac{2}{2011.\left(2011+2\right)}\right)\left(1\right)\)
Ta thấy:\(1-\frac{2}{n.\left(n+1\right)}=\frac{n.\left(n+1\right)-2}{n.\left(n+1\right)}=\frac{n^2+n-1-1}{n.\left(n+1\right)}=\frac{\left(n^2-1\right)+\left(n-1\right)}{n.\left(n+1\right)}=\frac{\left(n-1\right).\left(n+1\right)+\left(n-1\right).1}{n.\left(n+1\right)}=\frac{\left(n-1\right).\left(n+2\right)}{n.\left(n+1\right)}\left(2\right)\)
Thay (2) vào (1) ta có:
\(S=\frac{\left(2-1\right).\left(2+2\right)}{2.\left(2+1\right)}.\frac{\left(3-1\right).\left(3+2\right)}{3.\left(3+1\right)}...\frac{\left(2011-1\right).\left(2011+2\right)}{2011.\left(2011+1\right)}\)
=>\(S=\frac{1.4}{2.3}.\frac{2.5}{3.4}...\frac{2010.2013}{2011.2012}\)
=>\(S=\frac{\left(1.2....2010\right).\left(4.5...2013\right)}{\left(2.3...2011\right).\left(3.4...2012\right)}\)
=>\(S=\frac{1.2013}{2011.3}=\frac{2013}{6033}=\frac{671}{2011}\)
= \(\left(1-\frac{2}{2.3}\right).\left(1-\frac{2}{3.4}\right)...\left(1-\frac{1}{2011.2012}\right)=\frac{4}{2.3}.\frac{10}{3.4}.\frac{18}{4.5}...\frac{2010.2013}{2011.2012}\)
= \(\frac{\left(1.4\right).\left(2.5\right).\left(3.6\right)...\left(2010.2013\right)}{\left(2.3.4...2011\right).\left(3.4.5...2012\right)}=\frac{\left(1.2.3...2010\right).\left(4.5.6...2013\right)}{\left(2.3.4...2011\right).\left(3.4.5...2012\right)}=\frac{1.2013}{2011.3}=\frac{2013}{6033}\)
