\(C=\frac{1}{100}-\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
\(=\frac{1}{100}+\frac{1}{100}-\frac{1}{99}+\frac{1}{99}-\frac{1}{98}+\frac{1}{98}-\frac{1}{97}-...+\frac{1}{3}-\frac{1}{2}+\frac{1}{2}-1\)
\(=\frac{1}{100}+\frac{1}{100}-1\)
\(=-\frac{49}{50}\)
\(\Rightarrow50C=50.\frac{-49}{50}=-49\)