\(C=\frac{1}{100}-\frac{1}{100.99}-\frac{1}{99.98}-...-\frac{1}{2.1}\)
\(\Rightarrow C=\frac{1}{100}-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)\)
\(\Rightarrow C=\frac{1}{100}-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(\Rightarrow C=\frac{1}{100}-\left(1-\frac{1}{100}\right)\)
\(\Rightarrow C=\frac{1}{100}-1+\frac{1}{100}\)
\(\Rightarrow C=\frac{1}{50}-1\)
\(\Rightarrow50C=\left(\frac{1}{50}-1\right)50\)
\(\Rightarrow50C=1-50\)
\(\Rightarrow50C=-49\)
Vậy \(50C=-49\)