Ta có: 1/1.2 = 1/1 - 1/2 ; 1/2.3 = 1/2 - 1/3 ; 1/3.4 = 1/3 - 1/4 ; ...;1/99.100 = 1/99 - 1/100
Như vậy thì bài toán trên = 1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ...+ 1/99 - 1/100
Vậy tổng trên là:
1 - 1/100
= 99/100
Ta có:` 1/1.2 = 1/1 - 1/2 ; 1/2.3 = 1/2 - 1/3 ; 1/3.4 = 1/3 - 1/4 ; ...;1/99.100 = 1/99 - 1/100`
ta có :
`= 1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ...+ 1/99 - 1/100`
Vậy tổng trên là:
`1 - 1/100 = 99/100`
\(\dfrac{1}{2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}=\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}=\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{100}=1-\dfrac{1}{100}=\dfrac{99}{100}\)
\(\dfrac{1}{2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\)
\(=\dfrac{1}{2}+\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\right)\)
\(=\dfrac{1}{2}+\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+....+\dfrac{1}{99}-\dfrac{1}{100}\right)\)
\(=\dfrac{1}{2}+\left(\dfrac{1}{2}-\dfrac{1}{100}\right)\)
\(=\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{100}\)
\(=1-\dfrac{1}{100}\)
\(=\dfrac{99}{100}\)