Question

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

Answer 1

=> C = \(-\frac{1}{1.2}-\frac{1}{2.3}-...-\frac{1}{99.100}+\frac{1}{100}\)

=> C = \(-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)+\frac{1}{100}\)

=> C = \(-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)+\frac{1}{100}\)

=> C = \(-\left(1-\frac{1}{100}\right)+\frac{1}{100}\)

=>  C =\(-1+\frac{1}{100}+\frac{1}{100}\)

=> C = \(-1+\left(\frac{1}{100}+\frac{1}{100}\right)\)

=> C = \(-1+\frac{1}{50}\)

=> C =  \(-\frac{49}{50}\)

KL : C = \(-\frac{49}{50}\)