Question

\(D=\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot......\cdot\frac{99}{100}\) chứng minh rằng \(D< \frac{1}{10}\)

Answer 1

\(D=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{99}{100}\)

\(D< \frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{100}{101}\)

\(D^2< \frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}.\frac{5}{6}.\frac{6}{7}...\frac{99}{100}.\frac{100}{101}\)

\(D^2< \frac{1}{101}< \frac{1}{100}=\left(\frac{1}{10}\right)^2\)

=> \(D< \frac{1}{10}\left(đpcm\right)\)

Answer 2

\(D=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{99}{100}\)

\(D< \frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{100}{101}\)

\(D^2< \frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}.\frac{5}{6}.\frac{6}{7}...\frac{99}{100}.\frac{100}{101}\)

\(D^2< \frac{1}{101}< \frac{1}{100}=\left(\frac{1}{10}\right)^2\)

\(= >D< \frac{1}{10}\)

\(\text{k tui}\)