Question

Chứng minh rằng \(\frac{3}{1.4}+\frac{3}{2.6}+\frac{3}{3.8}+...+\frac{1}{2012.1342}\)<1.5

Answer 1

\(\frac{3}{1.4}+\frac{3}{2.6}+\frac{3}{3.8}+...+\frac{1}{2012.1342}\)

\(=\frac{3}{1.4}+\frac{3}{2.6}+\frac{3}{3.8}+...+\frac{3}{2012.4026}\)

\(=\frac{6}{2.4}+\frac{6}{4.6}+\frac{6}{4.8}+...+\frac{6}{4024.4026}\)

\(=3\cdot\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{4024.4026}\right)\)

\(=3\cdot\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{4024}-\frac{1}{4026}\right)\)

\(=3\cdot\left(\frac{1}{2}-\frac{1}{4026}\right)\)

\(=3\cdot\frac{1}{2}-3\cdot\frac{1}{4026}\)

\(=1,5-\frac{3}{4026}< 1,5\)