C= 1/100-(1/1.2+1/2.3+...+1/97.98+1/98.99+1/99.100)
C=1/100-(1-1/2+1/2-1/3+...+1/97-1/98+1/98-1/99+1/99-1/100)
C=1/100-(1-1/100)
C=1/100-99/100
C=-98/100=-49/50
\(C=\dfrac{1}{100}-\dfrac{1}{100.99}-\dfrac{1}{99.98}-\dfrac{1}{98.97}-...\dfrac{1}{3.2}-\dfrac{1}{2.1}\)
\(=-\left(\dfrac{1}{100.99}+\dfrac{1}{99.98}+\dfrac{1}{98.97}+...+\dfrac{1}{3.2}+\dfrac{1}{2.1}\right)+\dfrac{1}{100}\)
\(=-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{97.98}+\dfrac{1}{98.99}+\dfrac{1}{99.100}\right)+\dfrac{1}{100}\)
\(=-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{97}-\dfrac{1}{98}+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\right)+\dfrac{1}{100}\)
\(=-\left(1-\dfrac{1}{100}\right)+\dfrac{1}{100}\)
\(=\left(-1\right)+\dfrac{1}{50}=-\dfrac{49}{50}\)
theo bài ra ta có:
\(C=\dfrac{1}{100}-\dfrac{1}{100.99}-\dfrac{1}{99.98}-\dfrac{1}{98.97}-...-\dfrac{1}{3.2}-\dfrac{1}{2.1}\\ \Rightarrow C=\dfrac{1}{100}-\left(\dfrac{1}{100.99}+\dfrac{1}{99.98}+\dfrac{1}{98.97}+...+\dfrac{1}{3.2}+\dfrac{1}{2.1}\right)\\ \Rightarrow C=\dfrac{1}{100}-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{97.98}+\dfrac{1}{98.99}+\dfrac{1}{99.100}\right)\\ \Rightarrow C=\dfrac{1}{100}-\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{97}-\dfrac{1}{98}+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\right)\) \(\Rightarrow C=\dfrac{1}{100}-\left(\dfrac{1}{1}-\dfrac{1}{100}\right)\\ \Rightarrow C=\dfrac{1}{100}-\dfrac{99}{100}\\ \Rightarrow C=\dfrac{-98}{100}=\dfrac{-49}{50}\)
\(\dfrac{1}{100}\)\(-\)\(\dfrac{1}{100\times99}-\dfrac{1}{99\times98}-\dfrac{1}{98\times97}-...-\dfrac{1}{3\times2}-\dfrac{1}{2\times1}\)
\(\Rightarrow\dfrac{1}{100}-\left(\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+...+\dfrac{1}{98\times99}+\dfrac{1}{99\times100}\right)\)
\(\Rightarrow\dfrac{1}{100}-\left(1-\dfrac{1}{100}\right)\)
Vậy C=\(\dfrac{-49}{50}\)
