b) Ta có: \(B=\dfrac{1}{10\cdot11}+\dfrac{1}{11\cdot12}+\dfrac{1}{12\cdot13}+\dfrac{1}{13\cdot14}\)
\(=\dfrac{1}{10}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{14}\)
\(=\dfrac{1}{10}-\dfrac{1}{14}\)
\(=\dfrac{14}{140}-\dfrac{10}{140}\)
\(=\dfrac{4}{140}=\dfrac{1}{35}\)