Question

Cho \(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+................+\dfrac{1}{9^2}\)

Chứng minh \(\dfrac{2}{5}< A< \dfrac{8}{9}\)

Help me!!!!!!!!!!! tôi đang cần gấp!!!

Answer 1

câu này dễ.đầu óc phải linh hoat lên chứ cậu

Answer 2

Ta có : A=\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{9^2}\)

\(\Rightarrow A=\dfrac{1}{2.2}+\dfrac{1}{3.3}+\dfrac{1}{4.4}+...+\dfrac{1}{9.9}\)

^{\(\Rightarrow A=\dfrac{1}{2.2}+\dfrac{1}{3.3}+\dfrac{1}{4.4}+...+\dfrac{1}{9.9}\)}<\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{8.9}\)

\(\Rightarrow A< \dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{8}-\dfrac{1}{9}\)

\(\Rightarrow A< \dfrac{1}{1}-\dfrac{1}{9}\)

\(\Rightarrow A< \dfrac{9}{9}-\dfrac{1}{9}\)

\(\Rightarrow A< \dfrac{8}{9}\) (1)

\(\Rightarrow A=\dfrac{1}{2.2}+\dfrac{1}{3.3}+\dfrac{1}{4.4}+...+\dfrac{1}{9.9}\)>\(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{9.10}\)

\(\Rightarrow A>\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{1}{10}\)

\(\Rightarrow A>\dfrac{1}{2}-\dfrac{1}{10}\)

\(\Rightarrow A>\dfrac{5}{10}-\dfrac{1}{10}\)

\(\Rightarrow A>\dfrac{4}{10}\)

\(\Rightarrow A>\dfrac{2}{5}\) (2)

Từ (1) và (2)\(\Rightarrow\dfrac{2}{5}< A< \dfrac{8}{9}\)