Question

[Trang Ha]

A=2/3^2 + 2/5^2 + 2/7^2 + .... _ 2/99^2 

Chứng minh: 98/303<A<98/99 

Answer 1

[Nguyễn Lê Phước Thịnh]

\(\dfrac{2}{3^2}< \dfrac{2}{1\cdot3}=1-\dfrac{1}{3}\)

\(\dfrac{2}{5^2}< \dfrac{2}{3\cdot5}=\dfrac{1}{3}-\dfrac{1}{5}\)

...

\(\dfrac{2}{99^2}< \dfrac{2}{97\cdot99}=\dfrac{1}{97}-\dfrac{1}{99}\)

Do đó: \(A=\dfrac{2}{3^2}+\dfrac{2}{5^2}+...+\dfrac{2}{99^2}< 1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{97}-\dfrac{1}{99}\)=>\(A< 1-\dfrac{1}{99}=\dfrac{98}{99}\)

\(\dfrac{2}{3^2}>\dfrac{2}{3\cdot5}=\dfrac{1}{3}-\dfrac{1}{5}\)

\(\dfrac{2}{5^2}>\dfrac{2}{5\cdot7}=\dfrac{1}{5}-\dfrac{1}{7}\)

...

\(\dfrac{2}{99^2}>\dfrac{2}{99\cdot101}=\dfrac{1}{99}-\dfrac{1}{101}\)
Do đó: \(A=\dfrac{2}{3^2}+\dfrac{2}{5^2}+...+\dfrac{2}{99^2}>\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}\)

=>\(A>\dfrac{1}{3}-\dfrac{1}{101}=\dfrac{98}{303}\)

Do đó: \(\dfrac{98}{303}< A< \dfrac{98}{99}\)