Question

Cho \(P=\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{2019.2021}\)

chứng tỏ rằng P<1

Answer 1

Ta có : \(P=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2019}-\dfrac{1}{2020}=1-\dfrac{1}{2020}=\dfrac{2019}{2020}\)

mà \(2019< 2020\)nên P < 1 ( đpcm ) 

Answer 2

\(P=\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{2019.2021}\) 

\(P=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2019}-\dfrac{1}{2021}\) 

\(P=1-\dfrac{1}{2021}\) 

\(P=\dfrac{2020}{2021}\)

Vì \(\dfrac{2020}{2021}< 1\) ⇒ \(P< 1\) ( điều phải chứng minh )