Đặt
A = \(\dfrac{1}{3^2}+\dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{99^2}\)
= \(\dfrac{1}{3.3}+\dfrac{1}{4.4}+\dfrac{1}{5.5}+...+\dfrac{1}{99.99}\)
Vì \(\dfrac{1}{3.3}< \dfrac{1}{2.3}\)
\(\dfrac{1}{4.4}< \dfrac{1}{3.4}\)
................
\(\dfrac{1}{99.99}< \dfrac{1}{98.99}\)
=> A < \(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{98.99}\)
A < \(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{98}-\dfrac{1}{99}\)
A < \(\dfrac{1}{2}-\dfrac{1}{99}\)
A < \(\dfrac{97}{198}< \dfrac{99}{198}\)=\(\dfrac{1}{2}\)
=> A < \(\dfrac{1}{2}\)
=> \(\dfrac{1}{3^2}+\dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{99^2}\)< \(\dfrac{1}{2}\) < đpcm>
\(\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{99^2}< \dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{98.99}\)
\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{98}-\dfrac{1}{99}\)
\(=\dfrac{1}{2}-\dfrac{1}{99}< \dfrac{1}{2}\)
\(\Rightarrow\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{99^2}< \dfrac{1}{2}\left(đpcm\right)\)
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