Sửa đề: \(\dfrac{3}{1\cdot3}+\dfrac{3}{3\cdot5}+\dfrac{3}{5\cdot7}+...+\dfrac{3}{99\cdot101}\)
Ta có: \(\dfrac{3}{1\cdot3}+\dfrac{3}{3\cdot5}+\dfrac{3}{5\cdot7}+...+\dfrac{3}{99\cdot101}\)
\(=\dfrac{3}{2}\cdot\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{99\cdot101}\right)\)
\(=\dfrac{3}{2}\cdot\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)
\(=\dfrac{3}{2}\cdot\left(1-\dfrac{1}{101}\right)\)
\(=\dfrac{3}{2}\cdot\dfrac{100}{101}\)
\(=\dfrac{300}{202}=\dfrac{150}{101}\)
3/1.3+3/3.5+3/5.7+.....+3/99.100
=3.(1.3+1/3.5+...+1/99.100)
=3.1/2.(2/1.3+2/3.5+...=2/99.100)
=> ĐỀ SAI
