Question

cho A= \(\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+...+\dfrac{1}{99.100}\)

chứng minh rằng \(\dfrac{7}{12}< A< \dfrac{5}{6}\)

bạn nào làm xong trước mik tik

Answer 1

Ta có: \(A=\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+...+\dfrac{1}{99.100}\)

\(\Rightarrow A>\dfrac{1}{1.2}+\dfrac{1}{3.4}=\dfrac{1}{2}+\dfrac{1}{12}=\dfrac{14}{24}=\dfrac{7}{12}\)\(\left(1\right)\)

Lại có: \(A=\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+...+\dfrac{1}{99.100}\)

\(A=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

\(A=\left(1-\dfrac{1}{2}+\dfrac{1}{3}\right)-\left(\dfrac{1}{4}-\dfrac{1}{5}\right)-\left(\dfrac{1}{6}-\dfrac{1}{7}\right)-...-\left(\dfrac{1}{98}-\dfrac{1}{99}\right)-\dfrac{1}{100}\)

\(\Rightarrow A< 1-\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{5}{6}\)\(\left(2\right)\)

Từ \(\left(1\right)\) và \(\left(2\right)\) \(\Rightarrow\dfrac{7}{12}< A< \dfrac{5}{6}\)

Vậy \(\dfrac{7}{12}< A< \dfrac{5}{6}\) ( Điều phải chứng minh ).

Answer 2

Ta có:

\(A=\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+...+\dfrac{1}{99.100}\\ A=\left(\dfrac{1}{1.2}+\dfrac{1}{3.4}\right)+\dfrac{1}{5.6}+...+\dfrac{1}{99.100}\\ A=\dfrac{7}{12}+\dfrac{1}{5.6}+...+\dfrac{1}{99.100}>\dfrac{7}{12}\left(1\right)\\ \Rightarrow A>\dfrac{7}{12}\left(1\right)\)

Ta lại có:

\(A=\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+...+\dfrac{1}{99.100}\\ A=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{99}-\dfrac{1}{100}\\ A=\left(1-\dfrac{1}{2}+\dfrac{1}{3}\right)-\left(\dfrac{1}{4}-\dfrac{1}{5}\right)-...\left(\dfrac{1}{98}-\dfrac{1}{99}\right)-\dfrac{1}{100}\\ A=\dfrac{5}{6}-\left(\dfrac{1}{4}-\dfrac{1}{5}\right)-...\left(\dfrac{1}{98}-\dfrac{1}{99}\right)-\dfrac{1}{100}< \dfrac{5}{6}\\ \Rightarrow A=< \dfrac{5}{6}\left(2\right)\)

Từ (1) và (2) suy ra: \(\dfrac{7}{12}< A< \dfrac{5}{6}\left(dpcm\right)\)