Đặt \(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{23.24.25}\)
\(\Rightarrow2A=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{23.24.25}\)
\(=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+....+\frac{1}{23.24}-\frac{1}{24.25}\)
\(=\frac{1}{1.2}-\left(\frac{1}{2.3}-\frac{1}{2.3}\right)-\left(\frac{1}{3.4}-\frac{1}{3.4}\right)-\left(\frac{1}{4.5}-\frac{1}{4.5}\right)-....-\left(\frac{1}{23.24}-\frac{1}{23.24}\right)-\frac{1}{24.25}\)
\(=\frac{1}{1.2}-\frac{1}{24.25}\)
\(=\frac{1}{2}-\frac{1}{600}=\frac{299}{600}\)
Vậy : \(A=\frac{299}{600}\)
