Question

\(ChoE=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+...+\frac{1}{18\cdot19\cdot20}\)

Chứng minh rằng E<\(\frac{1}{4}\)

 

Answer 1

\(E=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{18.19.20}\)

\(E=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{18.19}-\frac{1}{19.20}\right)\)

\(E=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{19.20}\right)=\frac{1}{4}-\frac{1}{2.19.20}< \frac{1}{4}\left(đpcm\right)\)

Answer 2

Ta có : 

\(E=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{18.19.20}=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{18.19}-\frac{1}{19.20}\)

\(E=\frac{1}{2}-\frac{1}{380}=\frac{189}{380}< \frac{95}{380}=\frac{1}{4}\)

Vậy \(E< \frac{1}{4}\)

Chúc bạn học tốt ~

Answer 3

Àk nhầm rùi : 

Ta có : 

\(E=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{18.19.20}\)

\(\Leftrightarrow\)\(E=\frac{1}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{18.19.20}\right)\)

\(\Leftrightarrow\)\(E=\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{18.19}-\frac{1}{19.20}\right)\)

\(\Leftrightarrow\)\(E=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{19.20}\right)=\frac{1}{4}-\frac{1}{2.19.20}< \frac{1}{4}\)

Vậy \(E< \frac{1}{4}\)

Học tốt :)