Question

Chứng minh rằng

a) \(\frac{1}{5}<\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{98}-\frac{1}{99}<\frac{2}{5}\)

b) \(\frac{1}{15}<\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{99}{100}<\frac{1}{10}\)

 

Answer 1

OK. Tối nhớ giải hộ mik nha

Mik hứa sẽ lik-e cho bạn

 

Answer 2

mình ko biết

Answer 3

\(\left(1\right)\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{98}-\frac{1}{99}>\frac{1}{5}\)

\(=\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}\right)+\left(\frac{1}{6}-\frac{1}{7}\right)+\left(\frac{1}{8}-\frac{1}{9}\right)+...+\left(\frac{1}{98}-\frac{1}{99}\right)\)

\(=\frac{13}{60}+\left(\frac{1}{6}-\frac{1}{7}\right)+\left(\frac{1}{8}-\frac{1}{9}\right)+...\left(\frac{1}{98}-\frac{1}{99}\right)\)

Ta thấy        \(\frac{13}{60}>\frac{12}{60}=\frac{1}{5}\)

          \(\frac{1}{6}-\frac{1}{7}>0\)

          \(\frac{1}{8}-\frac{1}{9}>0\)

\(...\)\(\frac{1}{98}-\frac{1}{99}>0\)

\(\Rightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{98}-\frac{1}{99}>\frac{1}{5}\)

\(\left(2\right)\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{98}-\frac{1}{99}< \frac{2}{5}\)

\(=\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+\frac{1}{6}\right)-\left(\frac{1}{7}-\frac{1}{8}\right)-\left(\frac{1}{9}-\frac{1}{10}\right)-...-\left(\frac{1}{97}-\frac{1}{98}\right)-\frac{1}{99}\)

\(=\frac{23}{60}-\left(\frac{1}{7}-\frac{1}{8}\right)-\left(\frac{1}{9}-\frac{1}{10}\right)-...-\left(\frac{1}{97}-\frac{1}{98}\right)-\frac{1}{99}\)

Ta thấy   \(\frac{23}{60}< \frac{24}{60}=\frac{2}{5}\)

      \(\frac{1}{7}-\frac{1}{8}>0\)

     \(\frac{1}{9}-\frac{1}{10}>0\)

\(...\frac{1}{97}-\frac{1}{98}>0\)

                \(\frac{1}{99}>0\)

\(\Rightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{98}-\frac{1}{99}< \frac{2}{5}\)