Question

Chứng minh

\(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+...+\dfrac{1}{199\cdot200}< \dfrac{1}{2}\)

 

Answer 1

Ta có: \(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+...+\dfrac{1}{199\cdot200}\)

\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{199}-\dfrac{1}{200}\)

\(=\dfrac{1}{2}-\dfrac{1}{200}\)

\(=\dfrac{100-1}{200}=\dfrac{99}{200}< \dfrac{100}{200}=\dfrac{1}{2}\)(đpcm)

Answer 2

Ta có: \(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+...+\dfrac{1}{199\cdot200}\)

          \(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\cdot\cdot\cdot+\dfrac{1}{199}-\dfrac{1}{200}\)

          \(=\dfrac{1}{2}-\dfrac{1}{200}< \dfrac{1}{2}\)  (Đpcm)