Question

Cho A = \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{9^2}.\)Chứng minh \(\frac{2}{5}\)< A < \(\frac{8}{9}\)

Answer 1

\(A< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{8.9}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{8}-\frac{1}{9}=1-\frac{1}{9}=\frac{8}{9}\)

\(A>\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}=\frac{1}{2}-\frac{1}{10}=\frac{2}{5}\)

\(\Rightarrow\frac{2}{5}< A< \frac{8}{8}\)

Vây:.....

- Hok tốt ~

Answer 2

- k bn ơi ~

Answer 3

 A\(=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{8.9}.\)

\(=>\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}\)

=>\(1-\frac{1}{9}=\frac{8}{9}\)

=>  A  < 8/9

+) A\(=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{ }{9^2}>\) 1 /2.3+  1/3.4 +  1/4.5 + ... +  1/9.10

Tg tự nhé