Question

c/m:

\(1-\dfrac{1}{2^2}-\dfrac{1}{3^2}-\dfrac{1}{4^2}-..-\dfrac{1}{2004^2}>\dfrac{1}{2004}\)

Answer 1

Đặt \(A=1-\dfrac{1}{2^2}-\dfrac{1}{3^2}-\dfrac{1}{4^2}-...-\dfrac{1}{2004^2}\)

\(A=1-\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{2004^2}\right)\)

Đặt \(B=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{2004^2}\)

\(B=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{2004^2}< \dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{2003\cdot2004}\)

\(B< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2003}-\dfrac{1}{2004}\)

\(B< 1-\dfrac{1}{2004}\)

\(\Rightarrow B< \dfrac{2003}{2004}\)

\(\Rightarrow1-B>1-\dfrac{2003}{2004}\)

\(\Rightarrow A>\dfrac{1}{2004}\left(đpcm\right)\)