Question

Mọi người giải chi tiết giúp em với

Answer 1

Answer 2

\(\Leftrightarrow\dfrac{1}{3^2}+\dfrac{1}{4^2}+....+\dfrac{1}{100^2}< \dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\\ \Leftrightarrow Ta.thay:\\ \dfrac{1}{2.3}=\dfrac{1}{2}-\dfrac{1}{3};\dfrac{1}{3.4}=\dfrac{1}{3}-\dfrac{1}{4};...;\dfrac{1}{99.100}=\dfrac{1}{99}-\dfrac{1}{100}\\ \Leftrightarrow A< \dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\\ \Leftrightarrow A< \dfrac{1}{2}-\dfrac{1}{100}< \dfrac{1}{2}\\ \Rightarrow A< \dfrac{1}{2}\\ \Rightarrow\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}< \dfrac{1}{2}\left(đpcm\right)\)