Question

Cho A = \(\dfrac{1}{4}\)+\(\dfrac{1}{9}\)+\(\dfrac{1}{16}\)+....+\(\dfrac{1}{81}\)+\(\dfrac{1}{100}\) chứng tỏ rằng A > \(\dfrac{65}{132}\)

Answer 1

Có:

\(A=\dfrac{1}{4}+\dfrac{1}{9}+\dfrac{1}{16}+...+\dfrac{1}{81}+\dfrac{1}{100}\)

\(A=\dfrac{1}{4}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{9^2}+\dfrac{1}{10^2}\)

Mà: \(\dfrac{1}{3^2}>\dfrac{1}{3.4}\)

\(\dfrac{1}{4^2}>\dfrac{1}{4.5}\)

...

\(\dfrac{1}{9^2}>\dfrac{1}{9.10}\)

\(\dfrac{1}{10^2}>\dfrac{1}{10.11}\)

\(\Rightarrow A>\dfrac{1}{4}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{9.10}+\dfrac{1}{10.11}\)

\(A>\dfrac{1}{4}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{9.10}+\dfrac{1}{10.11}\)

\(\Rightarrow A>\dfrac{1}{4}+\dfrac{1}{3}-0-0-...-0-\dfrac{1}{11}\)

\(\Rightarrow A>\dfrac{1}{4}+\dfrac{1}{3}-\dfrac{1}{11}\)

\(\Rightarrow A>\dfrac{65}{132}\)

Chúc bạn học tốt!