Question

Chứng Minh Rằng A=\(\dfrac{1}{1+3}+\dfrac{1}{1+3+5}+\dfrac{1}{1+3+5+7}+...+\dfrac{1}{1+3+5+...+2017}\)\(< \dfrac{3}{4}\)

Answer 1

\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{1009^2}\)

Ta có: \(\dfrac{1}{2^2}=\dfrac{1}{4};\dfrac{1}{3^2}< \dfrac{1}{2.3};\dfrac{1}{4^2}< \dfrac{1}{3.4};...;\dfrac{1}{1009^2}< \dfrac{1}{1008.1009}\)

\(\Rightarrow\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{1009^2}< \dfrac{1}{4}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+.....+\dfrac{1}{1008.1009}\)\(\Rightarrow A< \dfrac{1}{4}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{1008}-\dfrac{1}{1009}\)

\(\Rightarrow A< \dfrac{1}{4}+\dfrac{1}{2}-\dfrac{1}{1009}\)

\(\Rightarrow A< \dfrac{3}{4}-\dfrac{1}{1009}\)

\(\Rightarrow A< \dfrac{3}{4}\left(đpcm\right)\)