\(\frac{1}{99}-\frac{1}{99.98}-\frac{1}{98.97}-\frac{1}{97.96}+......+\frac{1}{2.1}\)
\(= \frac{1}{99}-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{98.99}\right)\)
\(= \frac{1}{99}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{98}-\frac{1}{99}\right)\)
\(= \frac{1}{99}-\left(1-\frac{1}{99}\right)\)
\(= \frac{1}{99}-\frac{98}{99}\)
\(= \frac{-97}{99}\)
\(\dfrac{-1}{100\cdot99}+\dfrac{-1}{99\cdot98}+\dfrac{-1}{98\cdot97}+...+\dfrac{-1}{3\cdot2}+\dfrac{-1}{2\cdot1}\\ \left(-1\right)\cdot\left(\dfrac{1}{100\cdot99}+\dfrac{1}{99\cdot98}+\dfrac{1}{98\cdot97}+...+\dfrac{1}{3\cdot2}+\dfrac{1}{2\cdot1}\right)\\ =\left(-1\right)\cdot\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{98\cdot99}+\dfrac{1}{99\cdot100}\right)\\ =\left(-1\right)\cdot\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\right)\\ =\left(-1\right)\cdot\left(1-\dfrac{1}{100}\right)\\ =\left(-1\right)\cdot\dfrac{99}{100}\\ =\dfrac{-99}{100}\)