\(K=1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+....+\frac{1}{45}\)
\(K=1+\frac{1}{\frac{2\cdot3}{2}}+\frac{1}{\frac{3\cdot4}{2}}+\frac{1}{\frac{4\cdot5}{2}}+....+\frac{1}{\frac{9\cdot10}{2}}\)
\(K=1+\left[2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{9}-\frac{1}{10}\right)\right]\)
\(K=1+\left[2\left(\frac{1}{2}-\frac{1}{10}\right)\right]\)
\(K=1+\left(2\cdot\frac{2}{5}\right)=1+\frac{4}{5}=\frac{9}{5}\)
Vì \(\frac{9}{5}< 2\)\(\Rightarrow K< 2\)
\(K=1+\frac{2}{6}+\frac{2}{12}+...+\frac{2}{90}\)
\(K=1+\left(\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(K=1+\left(\frac{1}{2}-\frac{1}{10}\right)\)
\(K=1+\frac{2}{5}\)
\(K=\frac{7}{5}\)
Ta có ; K = \(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+.....+\frac{1}{45}\)
\(=1+\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+.....+\frac{2}{90}\)
\(=1+2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+.....+\frac{1}{90}\right)\)
\(=1+2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+.....+\frac{1}{9.10}\right)\)
\(=1+2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{9}-\frac{1}{10}\right)\)
\(=1+2\left(\frac{1}{2}-\frac{1}{100}\right)\)
\(=1+\left(1-\frac{1}{50}\right)\)
\(=1+\frac{49}{50}=\frac{99}{50}\)
\(K=1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{45}\)
\(K=\frac{2}{\frac{2\cdot3}{2}}+\frac{2}{\frac{3\cdot4}{2}}+\frac{2}{\frac{4\cdot5}{2}}+....+\frac{2}{\frac{9\cdot10}{2}}\)
\(K=2\cdot\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{9}-\frac{1}{10}\right)\)
\(K=2\cdot\left(\frac{1}{2}-\frac{1}{10}\right)=2\cdot\frac{2}{5}=\frac{4}{5}\)
Ta thấy \(\frac{4}{5}< 2\)\(\Rightarrow K< 2\)