Question

[Lãnh Hàn Thiên Băng]

Chứng minh

 A= \(\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+.....+\frac{1}{199}+\frac{1}{100}< 1\) \(1\)

Answer 1

[Kiệt Nguyễn]

                                  Giải

\(A=\frac{1}{100}+\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{199}\)

\(\Rightarrow A< \frac{1}{100}+\frac{1}{100}+\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}\)

\(\Rightarrow A< \frac{100}{100}=1\)

Vậy A < 1 (đpcm)