Question

Cho \(A=\dfrac{2}{3^2}+\dfrac{2}{5^2}+\dfrac{2}{7^2}+.............+\dfrac{2}{2007^2}\)

Chứng minh \(A< \dfrac{1003}{2008}\)

Help me!!!!!!!!!!

Answer 1

Xét p/s A=\(\dfrac{2}{3^2}+\dfrac{2}{5^2}+...........+\dfrac{2}{2007^2}\)

A<\(\dfrac{2}{2.4}+\dfrac{2}{4.6}+...........+\dfrac{2}{2006.2008}\)

A<\(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{2006}-\dfrac{1}{2008}\)

A<\(\dfrac{1}{2}-\dfrac{1}{2008}\)

A<\(\dfrac{1003}{2008}\)

Ta có đpcm

Answer 2

Ta thấy với k \(\in\) N* thì k² > (k - 1)(k + 1).

Thật vậy, ta có (k - 1)(k + 1) = k(k + 1) - (k + 1) = k² + k - k - 1 = k² - 1 < k².

Từ đó suy ra: 3² > 2 . 4; 5² > 4 . 6; 7² > 6 . 8;...; 2007² > 2006 . 2008.

\(\Rightarrow\dfrac{2}{3^2}< \dfrac{2}{2.4};\dfrac{2}{5^2}< \dfrac{2}{4.6};\dfrac{2}{7^2}< \dfrac{2}{6.8};...;\dfrac{2}{2007^2}< \dfrac{2}{2006.2008}\)

\(\Rightarrow A< \dfrac{2}{2.4}+\dfrac{2}{4.6}+\dfrac{2}{6.8}+...+\dfrac{2}{2006.2008}\)

\(\Rightarrow A< \dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{2006}-\dfrac{1}{2008}\)

\(\Rightarrow A< \dfrac{1}{2}-\dfrac{1}{2008}=\dfrac{1003}{2008}\)