Question

Chứng minh \(S=\frac{1}{50}+\frac{1}{51}+...+\frac{1}{98}+\frac{1}{99}>\frac{1}{2}\)

 

Answer 1

Ta có:

\(\frac{1}{2}=\frac{1}{100}+\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}\)(50 PS)

Vì \(\frac{1}{50}>\frac{1}{100}\)

     \(\frac{1}{51}>\frac{1}{100}\)

      \(\frac{1}{52}>\frac{1}{100}\)

       .......................

      \(\frac{1}{99}>\frac{1}{100}\)

\(=>\left(\frac{1}{50}+\frac{1}{51}+\frac{1}{52}+...+\frac{1}{99}\right)>\left(\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}\right)\)                                                                                                    ( có 50 PS)

\(=>\frac{1}{50}+\frac{1}{51}+\frac{1}{52}+...+\frac{1}{99}>\frac{1}{2}\)

Answer 2

Tổng S có : (99 - 50) : 1 + 1 = 50 (số)\(S=\frac{1}{50}+\frac{1}{51}+....+\frac{1}{98}+\frac{1}{99}>\frac{1}{100}+\frac{1}{100}+....+\frac{1}{100}+\frac{1}{100}\)(50 phân số \(\frac{1}{5}\)) = \(\frac{1}{100}.50=\frac{1}{2}\)

Vậy S > \(\frac{1}{2}\)