\(C=\frac{1}{100}-\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
\(C=\frac{1}{100}-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{98.99}+\frac{1}{99.100}\right)\)
\(C=\frac{1}{100}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)
\(C=\frac{1}{100}-\left(1-\frac{1}{100}\right)\)
\(C=\frac{1}{100}-1+\frac{1}{100}\)
\(C=\frac{-49}{50}\)
C = 1/100 - 1/100.99 - 1/99.98 - 1/98.97 - ... - 1/3.2 - 1/2.1
C = 1/100 - (1/100.99 + 1/99.98 + 1/98.97 + ... + 1/3.2 + 1/2.1)
C = 1/100 - (1/1.2 + 1/2.3 + ... + 1/98.99 + 1/99.100)
C = 1/100 - (1 - 1/2 + 1/2 - 1/3 + ... + 1/98 - 1/99 + 1/99 - 1/100)
C = 1/100 - (1 - 1/100)
C = 1/100 - 99/100
C = -98/100 = -49/50
\(c=\frac{1}{100}-\frac{1}{100.98}\frac{1}{99.98}\frac{1}{98.97}-......-\frac{1}{3.2}-\frac{1}{2.1}\)=\(\frac{1}{100}-\left[\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}+\frac{1}{99.100}\right]\) =\(\frac{1}{100}-\left[1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{98}+\frac{1}{99}-\frac{1}{100}\right]\)=\(\frac{1}{100}-\left[1-\frac{1}{100}\right]=\frac{1}{100}-\frac{99}{100}=\frac{-98}{100}=\frac{49}{50}\)
