Ta có: \(A=\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+...+\dfrac{2}{99\cdot101}\)
\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{99}-\dfrac{1}{101}\)
\(=\dfrac{1}{3}-\dfrac{1}{101}\)
\(=\dfrac{98}{303}\)
Với công thức \(\dfrac{a}{x.\left(x+a\right)}=\dfrac{a}{x}-\dfrac{a}{x+a}\)
Ta có: \(A=\dfrac{2}{3}-\dfrac{2}{5}+\dfrac{2}{5}-\dfrac{2}{7}+...+\dfrac{2}{99}-\dfrac{2}{101}\)
\(=\dfrac{2}{3}-\dfrac{2}{101}\)