\(\left(1-\dfrac{2}{2.3}\right)\left(1-\dfrac{2}{3.4}\right)...\left(1-\dfrac{2}{99.100}\right)\)
\(=\left(\dfrac{2.3}{2.3}-\dfrac{2}{2.3}\right)\left(\dfrac{3.4}{3.4}-\dfrac{2}{3.4}\right)...\left(\dfrac{99.100}{99.100}-\dfrac{2}{99.100}\right)\)
\(=\dfrac{4}{2.3}.\dfrac{10}{3.4}...\dfrac{9898}{99.100}\)
\(=\dfrac{1.4}{2.3}.\dfrac{2.5}{3.4}...\dfrac{98.101}{99.100}\)
\(=\dfrac{1.2...98}{2.3...99}.\dfrac{4.5...101}{3.4...100}=\dfrac{1}{99}.\dfrac{101}{3}=\dfrac{101}{297}\)
Dạng tổng quát với n = 2;3;...;99
\(1-\dfrac{2}{n\left(n+1\right)}=\dfrac{n^2+n-2}{n\left(n+1\right)}=\dfrac{\left(n+2\right)\left(n-1\right)}{n\left(n+1\right)}\)
Ta có : \(B=\dfrac{4.1}{2.3}.\dfrac{5.2}{3.4}.\dfrac{6.3}{4.5}...\dfrac{100.97}{98.99}.\dfrac{101.98}{99.100}\)
\(=\dfrac{\left(4.5.6...101\right).\left(1.2.3...98\right)}{\left(2.3.4...99\right)\left(3.4.5...100\right)}\) = \(\dfrac{100.101}{2.3}.\dfrac{2}{99.100}\)
= \(\dfrac{101}{3.99}=\dfrac{101}{297}\)
