Question

Tính S gồm 23 số hạng:  \(S=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{\left(n-1\right).n.\left(n+1\right)}+...+\frac{1}{23.24.25}\)

Answer 1

\(S=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{\left(n-1\right).n.\left(n+1\right)}+...+\frac{1}{23.24.25}\)

\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{\left(n-1\right).n}-\frac{1}{n.\left(n+1\right)}+...+\frac{1}{23.24}-\frac{1}{24.25}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{24.25}\right)=\frac{299}{1200}\)

Answer 2

\(=\frac{1}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{23.24.25}\right)=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{23.24}-\frac{1}{24.25}\right)\)

\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{600}\right)=\frac{1}{2}.\frac{299}{600}=\frac{299}{1200}\)