Question

Cho \(B=\dfrac{1}{5^2}+\dfrac{1}{6^2}+.......+\dfrac{1}{100^2}\)

C/m: \(\dfrac{1}{6}< B< \dfrac{1}{4}\)

Answer 1

Ta có: \(B=\dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}< \dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{99.100}\)

\(=\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

\(=\dfrac{1}{4}-\dfrac{1}{100}< \dfrac{1}{4}\)

\(B=\dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}>\dfrac{1}{5^2}+\dfrac{1}{6.7}+...+\dfrac{1}{100.101}\)

\(=\dfrac{1}{5^2}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{100}-\dfrac{1}{101}\)

\(=\dfrac{1}{5^2}+\dfrac{1}{6}-\dfrac{1}{101}>\dfrac{1}{6}\left(\dfrac{1}{5^2}>\dfrac{1}{101}\right)\)

\(\Rightarrow\dfrac{1}{6}< B< \dfrac{1}{4}\)