Question

Cho B = \(\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}+...+\dfrac{1}{120}\)

Chứng minh B < \(\dfrac{1}{2}\)

Answer 1

\(B=\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}+...+\dfrac{1}{120}\)

\(\Leftrightarrow B=\dfrac{2}{20}+\dfrac{2}{30}+\dfrac{2}{42}+...+\dfrac{2}{240}\)

\(\Leftrightarrow B=2\left(\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+...+\dfrac{1}{240}\right)\)

\(\Leftrightarrow B=2\left(\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{15.16}\right)\)

\(\Leftrightarrow B=2\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{15}-\dfrac{1}{16}\right)\)

\(\Leftrightarrow B=2\left(\dfrac{1}{4}-\dfrac{1}{16}\right)=\dfrac{3}{8}\)

Vì \(\dfrac{3}{8}< \dfrac{1}{2}\)

\(\Rightarrow B< \dfrac{1}{2}\left(ĐPCM\right)\)