Question

[Khuất Đăng Dương]

chứng minh rằng:1<1:5+1:6+1:7+......+1:17<2

Answer 1

[Đoàn Đức Hà]

\(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+...+\frac{1}{17}\)

\(=\left(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}\right)+\left(\frac{1}{9}+\frac{1}{10}+...+\frac{1}{17}\right)\)

\(>\left(\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}\right)+\left(\frac{1}{18}+\frac{1}{18}+...+\frac{1}{18}\right)\)(\(9\)số hạng \(\frac{1}{18}\))

\(=\frac{4}{8}+\frac{9}{18}=1\). 

\(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+...+\frac{1}{17}\)

\(=\left(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}\right)+\left(\frac{1}{9}+\frac{1}{10}+...+\frac{1}{17}\right)\)

\(< \left(\frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}\right)+\left(\frac{1}{9}+\frac{1}{9}+...+\frac{1}{9}\right)\)(\(9\)số hạng \(\frac{1}{9}\))

\(=\frac{4}{4}+\frac{9}{9}=2\)

Answer 2

[Tiêu Chiến]

Xét $\frac{1}{5}+\frac{1}{6}>\frac{1}{17}.6=\frac{6}{17}$

và $\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+...+\frac{1}{17}>\frac{1}{17}+\frac{1}{17}+\frac{1}{17}+...+\frac{1}{17}=\frac{1}{17}.11=\frac{11}{17}$

Do đó $\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+...+\frac{1}{17}>\frac{6}{17}+\frac{11}{17}=\frac{17}{17}=1$ (1)

Lại có $\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+...+\frac{1}{17}=\left(\frac{1}{5}+\frac{1}{6}+...+\frac{1}{10}\right)+\left(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{17}\right)$

$<\left(\frac{1}{10}+\frac{1}{10}+...+\frac{1}{10}\right)+\left(\frac{1}{17}+\frac{1}{17}+...+\frac{1}{17}\right)=\frac{1}{10}.6+\frac{1}{17}.7=1\frac{1}{85}<2$ (2)

Từ (1) và (2) suy ra điều phải chứng minh