Question

[Trần Kim Anh]

Cho M = \(\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+......+\frac{1}{1+2+3+.....+89}\)

Chứng minh M\(< \frac{2}{3}\)

Answer 1

[Nguyễn Ngọc Anh Minh]

\(M=\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+...+\frac{1}{4005}\)

\(\frac{M}{2}=\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{8010}\)

\(\frac{M}{2}=\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{89x90}\)

\(\frac{M}{2}=\frac{4-3}{3.4}+\frac{5-4}{4.5}+\frac{6-5}{5.6}+...+\frac{90-89}{89.90}\)

\(\frac{M}{2}=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{89}-\frac{1}{90}=\frac{1}{3}-\frac{1}{90}\)

\(M=\frac{2}{3}-\frac{2}{90}< \frac{2}{3}\)