\(B=\dfrac{3}{1.3}+\dfrac{3}{3.5}+...+\dfrac{3}{99.101}\)
\(=\dfrac{3}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{99.101}\right)\)
\(=\dfrac{3}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)
\(=\dfrac{3}{2}\left(1-\dfrac{1}{101}\right)\)
\(=\dfrac{3}{2}.\dfrac{100}{101}=\dfrac{300}{202}=\dfrac{50}{101}\)
Vậy \(B=\dfrac{50}{101}\)
\(B=\dfrac{3}{1\cdot3}+\dfrac{3}{3\cdot5}+...+\dfrac{3}{99\cdot101}\)
\(=\dfrac{3}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{99\cdot101}\right)\)
\(=\dfrac{3}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)
\(=\dfrac{3}{2}\left(1-\dfrac{1}{101}\right)\)
\(=\dfrac{3}{2}\left(\dfrac{101}{101}-\dfrac{1}{101}\right)\)
\(=\dfrac{3}{2}\cdot\dfrac{100}{101}\)
\(=\dfrac{150}{101}\)
