\(\dfrac{N}{2}=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{49\cdot50}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\\ =1-\dfrac{1}{50}< 1\\ N< 2\)
Ta có: \(N=\dfrac{2}{1\cdot2}+\dfrac{2}{2\cdot3}+\dfrac{2}{3\cdot4}+...+\dfrac{2}{49\cdot50}\)
\(=2\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{49\cdot50}\right)\)
\(=2\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)\)
\(=2\left(1-\dfrac{1}{50}\right)\)
\(=2\cdot\dfrac{49}{50}=\dfrac{49}{25}< \dfrac{50}{25}=2\)
Vậy: N<2
Ta có: =2(11⋅2+12⋅3+13⋅4+...+149⋅50)=2(11⋅2+12⋅3+13⋅4+...+149⋅50)
=2(1−150)=2(1−150)
