Question

Cho \(C=\frac{1}{100}-\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-...-\frac{1}{3.2}-\frac{1}{2.1}\).

Khi đó 5C=______

Answer 1

\(C=\frac{1}{100}-\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-...-\frac{1}{3.2}-\frac{1}{2.1}\)

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

\(=\frac{1}{100}-\frac{1}{99}+\frac{1}{100}-\frac{1}{98}+\frac{1}{99}-\frac{1}{97}+\frac{1}{98}-...-\frac{1}{2}+\frac{1}{3}-\frac{1}{1}+\frac{1}{2}\)

\(=\frac{1}{100}+\frac{1}{100}-1=\frac{1}{50}-\frac{50}{50}=-\frac{49}{50}\Rightarrow5C=-\frac{49}{10}\)