Question

CM\(\dfrac{1}{5}< \dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{100^2}< \dfrac{1}{3}\)

Answer 1

*Có: \(A=\dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{100^2}>\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{100.101}\)\(>\dfrac{1}{4}-\dfrac{1}{101}=\dfrac{97}{404}\)\(=\dfrac{970}{4040}\)

Có: \(\dfrac{1}{5}=\dfrac{808}{4040}\)

\(\Rightarrow\dfrac{1}{5}< A\)

*Có: \(A=\dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{100^2}< \dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{99.100}\)\(=\dfrac{1}{3}-\dfrac{1}{100}< \dfrac{1}{3}\)

\(\Rightarrow A< \dfrac{1}{3}\)

Vậy \(\dfrac{1}{5}< A< \dfrac{1}{3}\)