Đề sai nha:
Sửa lại:
Cho \(A=\dfrac{1}{4}+\dfrac{1}{9}+\dfrac{1}{16}+...+\dfrac{1}{81}+\dfrac{1}{100}\). Chứng tỏ rằng \(A>\dfrac{65}{132}\)
Giải:
Có:
\(A=\dfrac{1}{4}+\dfrac{1}{9}+\dfrac{1}{16}+...+\dfrac{1}{81}+\dfrac{1}{100}\)
\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{9^2}+\dfrac{1}{10^2}\)
Mà: \(\dfrac{1}{3^2}>\dfrac{1}{3.4}\);
\(\dfrac{1}{4^2}>\dfrac{1}{4.5}\);
...
\(\dfrac{1}{9^2}>\dfrac{1}{9.10}\);
\(\dfrac{1}{10^2}>\dfrac{1}{10.11}\).
\(\Rightarrow A>\dfrac{1}{2^2}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{9.10}+\dfrac{1}{10.11}\)
\(A>\dfrac{1}{2^2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{11}\)
\(\Rightarrow A>\dfrac{1}{2^2}+\dfrac{1}{3}-0-0-...-0-\dfrac{1}{11}\)
\(\Rightarrow A>\dfrac{1}{2^2}+\dfrac{1}{3}-\dfrac{1}{11}\)
\(\Rightarrow A>\dfrac{65}{132}\)
Chúc bạn học tốt!
