Đặt \(A=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{\left(n-1\right)\cdot n\cdot\left(n+1\right)}\)
Ta có: \(A=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{\left(n-1\right)\cdot n\cdot\left(n+1\right)}\)
\(\Rightarrow2\cdot A=\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+\frac{2}{3\cdot4\cdot5}+...+\frac{2}{\left(n-1\right)\cdot n\cdot\left(n+1\right)}\)
\(\Rightarrow2\cdot A=\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+\frac{1}{3\cdot4}-\frac{1}{4\cdot5}+...+\frac{1}{\left(n-1\right)\cdot n}-\frac{1}{n\left(n+1\right)}\)
\(\Rightarrow2\cdot A=\frac{1}{2}-\frac{1}{n\left(n+1\right)}\)
\(\Rightarrow2\cdot A=\frac{n\left(n+1\right)-2}{2n\left(n+1\right)}\)
hay \(A=\frac{n^2+n-2}{4n\left(n+1\right)}\)
