Question

Cho A=\(\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\frac{1}{5\cdot6}+......+\frac{1}{99\cdot100}\)CMR:\(\frac{7}{12}\)<A<\(\frac{5}{6}\)

Answer 1

ta có: A=\(\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}=>A=\left(\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+..+\frac{1}{75}\right)+\left(\frac{1}{76}+\frac{1}{77}+...+\frac{1}{100}\right)\)

\(\frac{1}{51}>\frac{1}{52}>\frac{1}{53}>...>\frac{1}{75};\frac{1}{76}>\frac{1}{77}>...>\frac{1}{100}\)

do đó:\(A>\frac{1}{75}.25+\frac{1}{100}.25=>A>\frac{1}{3}+\frac{1}{4}=\frac{7}{12}\)  (1)

lại có: \(A<\frac{1}{51}.25+\frac{1}{76}.25<\frac{1}{50}.25+\frac{1}{75}.25=\frac{1}{2}+\frac{1}{3}=\frac{5}{6}\)   (2)

từ (1) và (2)=>7/12<A<5/6(đpcm)