Question

 Tính

S=\(\frac{1}{1.3}\)​- \(\frac{1}{24}\) + \(\frac{1}{3.5}\)-\(\frac{1}{4.6}\)+...................+ \(\frac{1}{97.99}\)- \(\frac{1}{98.100}\)

( GIÚP MIK VS MIK ĐANG CẦN GẤP LẮM Ạ ! )

Answer 1

\(=\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{97.99}\right)-\left(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{98.100}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{98}-\frac{1}{100}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{99}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{100}\right)\)

\(=\frac{1}{2}\cdot\frac{98}{99}-\frac{1}{2}\cdot\frac{49}{100}\)

\(=\frac{1}{2}\left(\frac{98}{99}-\frac{49}{100}\right)=\frac{1}{2}\cdot\frac{4949}{9900}=\frac{4949}{19800}\)