Đặt \(A=\frac{1}{1.3}+\frac{1}{2.4}+\frac{1}{3.5}+...+\frac{1}{97.99}+\frac{1}{99.100}\)
\(\Rightarrow2A=2\left(\frac{1}{1.3}+\frac{1}{2.4}+\frac{1}{3.5}+...+\frac{1}{97.99}+\frac{1}{98.100}\right)\)
\(\Rightarrow2A=\frac{2}{1.3}+\frac{2}{2.4}+\frac{2}{3.5}+...+\frac{2}{97.99}+\frac{2}{98.100}\)
\(\Rightarrow2A=\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{97.99}\right)+\left(\frac{2}{2.4}+\frac{1}{4.6}+...+\frac{1}{98.100}\right)\)
\(\Rightarrow2A=\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\right)+\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{98}-\frac{1}{100}\right)\)
\(\Rightarrow2A=\left(1-\frac{1}{99}\right)+\left(\frac{1}{2}-\frac{1}{100}\right)\)
\(\Rightarrow2A=\left(\frac{99}{99}-\frac{1}{99}\right)+\left(\frac{50}{100}-\frac{1}{100}\right)\)
\(\Rightarrow2A=\frac{98}{99}+\frac{49}{100}=\frac{9800}{9900}+\frac{4851}{9900}=\frac{14651}{9900}\)
\(\Rightarrow A=\frac{14651}{9900}:2=\frac{14651}{9900}.\frac{1}{2}=\frac{14651}{19800}\)
bạn nhớ thử lại nhé :)
