a: \(A=\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{13\cdot15}\)
\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{13}-\dfrac{1}{15}\)
\(=\dfrac{1}{3}-\dfrac{1}{15}=\dfrac{4}{15}\)
b: \(B=\dfrac{6}{1\cdot4}+\dfrac{6}{4\cdot7}+...+\dfrac{6}{97\cdot100}\)
\(=2\left(\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+...+\dfrac{3}{97\cdot100}\right)\)
\(=2\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{100}\right)\)
\(=2\left(1-\dfrac{1}{100}\right)=2\cdot\dfrac{99}{100}=\dfrac{99}{50}\)
c: \(C=\dfrac{3}{3\cdot5}+\dfrac{3}{5\cdot7}+...+\dfrac{3}{99\cdot101}\)
\(=\dfrac{3}{2}\left(\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{99\cdot101}\right)\)
\(=\dfrac{3}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)
\(=\dfrac{3}{2}\left(\dfrac{1}{3}-\dfrac{1}{101}\right)=\dfrac{3}{2}\cdot\dfrac{98}{303}=\dfrac{49}{101}\)
