\(A=\dfrac{1}{1+3}+\dfrac{1}{1+3+5}+...+\dfrac{1}{1+3+...+2017}\)
\(=\dfrac{1}{4}+\dfrac{1}{9}+...+\dfrac{1}{1+3+...+2017}\)
\(=\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{3}\right)^2+...+\left(\dfrac{1}{1007}\right)^2\)
\(\Rightarrow A< \left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{1006\cdot1007}\right)\)
\(=\dfrac{1}{4}+\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{1006}-\dfrac{1}{1007}\right)\)
\(=\dfrac{1}{4}+\dfrac{1}{2}-\dfrac{1}{1007}< \dfrac{1}{4}+\dfrac{1}{2}=\dfrac{3}{4}\)
Vậy: \(A< \dfrac{3}{4}\)
Ta có: \(1+3+5+...+\left(2n-1\right)=n^2\)
\(\Rightarrow A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{1009^2}\)
\(\Rightarrow A< \dfrac{1}{2^2-1}+\dfrac{1}{3^2-1}+\dfrac{1}{4^2-1}+...+\dfrac{1}{1009^2-1}\)
\(\Rightarrow A< \dfrac{1}{1.3}+\dfrac{1}{2.4}+\dfrac{1}{3.5}+\dfrac{1}{4.6}+...+\dfrac{1}{1008.1010}\)
\(\Rightarrow A< \dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{1008}-\dfrac{1}{1010}\right)\)
\(\Rightarrow A< \dfrac{1}{2}\left(1+\dfrac{1}{2}-\dfrac{1}{1009}-\dfrac{1}{1010}\right)< \dfrac{1}{2}\left(1+\dfrac{1}{2}\right)\)
\(\Rightarrow A< \dfrac{3}{4}\)
Help me PLS
help me, pls. Sắp đến giờ nộp bài rồi, các bn giúp mình với