Question

[Nguyễn Hoành Nhật Minh]

A = 1/21+ 1/22 + 1/23+ 1/24 + ........+1/79 + 1/80

so sánh A với 39/ 40

Answer 1

[Nobi Nobita]

\(A=\frac{1}{21}+\frac{1}{22}+\frac{1}{23}+.....+\frac{1}{79}+\frac{1}{80}\)

\(=\left(\frac{1}{21}+\frac{1}{22}+.....+\frac{1}{50}\right)+\left(\frac{1}{51}+\frac{1}{52}+.....+\frac{1}{80}\right)\)

Ta có: \(21< 50\)\(\Rightarrow\frac{1}{21}>\frac{1}{50}\)

Tương tự ta có: \(\frac{1}{22}>\frac{1}{50}\); .......... ; \(\frac{1}{49}>\frac{1}{50}\)

Vì \(51< 80\)\(\Rightarrow\frac{1}{51}>\frac{1}{80}\)

Tương tự ta có: \(\frac{1}{52}>\frac{1}{80}\); ......... ; \(\frac{1}{79}>\frac{1}{80}\)

\(\Rightarrow A>\left(\frac{1}{50}+\frac{1}{50}+.....+\frac{1}{50}\right)+\left(\frac{1}{80}+\frac{1}{80}+....+\frac{1}{80}\right)\)

\(=30.\frac{1}{50}+30.\frac{1}{80}=\frac{3}{5}+\frac{3}{8}=\frac{39}{40}\)

Vậy \(A>\frac{39}{40}\)