Question

Cho \(S=\frac{1}{1.3.5}+\frac{1}{3.5.7}+\frac{1}{5.7.9}+...+\frac{1}{2003.2005.2007}\).

1) Tính gần đúng S

2) Tính đúng S (biểu diễn dưới dạng phân số).

Answer 1

1)\(S=\frac{1}{1.3.5}+\frac{1}{3.5.7}+\frac{1}{5.7.9}+...+\frac{1}{2003.2005.2007}\)

\(\Rightarrow4S=\frac{4}{1.3.5}+\frac{4}{3.5.7}+\frac{4}{5.7.9}+...+\frac{4}{2003.2005.2007}\)

\(=\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+\frac{1}{5.7}-\frac{1}{7.9}+...+\frac{1}{2003.2005}-\frac{1}{2005.2007}\)

\(=\frac{1}{3}-\frac{1}{4024035}=\frac{1341345}{4024035}=\frac{1}{3}\)

\(\Rightarrow S=\frac{1}{3}:4\approx0,08\)

2)\(S=\frac{1}{3}:4=\frac{1}{12}\)

Answer 2

\(S=\frac{1}{1.3.5}+\frac{1}{3.5.7}+\frac{1}{5.7.9}+...+\frac{1}{2003.2005.2007}\)

\(S=\frac{2}{2}.\frac{1}{1.3.5}+\frac{2}{2}.\frac{1}{3.5.7}+\frac{2}{2}.\frac{1}{5.7.9}+...+\frac{2}{2}.\frac{1}{2003.2005.2007}\)

\(S=\frac{1}{2}.\frac{2}{1.3.5}+\frac{1}{2}.\frac{2}{3.5.7}+\frac{1}{2}.\frac{2}{5.7.9}+...+\frac{1}{2}.\frac{2}{2003.2005.2007}\)

\(S=\frac{1}{2}.\left(\frac{2}{1.3.5}+\frac{2}{3.5.7}+\frac{2}{5.7.9}+...+\frac{2}{2003.2005.2007}\right)\)

\(S=\frac{1}{2}.\left(\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+\frac{1}{5.7}-\frac{1}{7.9}+...+\frac{1}{2003.2005}-\frac{1}{2005.2007}\right)\)

\(S=\frac{1}{2}\left(\frac{1}{1.3}-\frac{1}{2005.2007}\right)=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{4024035}\right)=\frac{1}{2}.\frac{1341345}{4024035}=\frac{1}{2}.\frac{1}{3}=\frac{1}{6}\)

Vậy \(S=\frac{1}{6}\)