A = \(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{48.50}\)
= \(\frac{1}{2}.\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{48.50}\right)\)
= \(\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{48}-\frac{1}{50}\right)\)
= \(\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{50}\right)\)
= \(\frac{1}{2}.\frac{12}{25}\)
= \(\frac{6}{25}\)
\(A=\frac{1}{2\cdot4}+\frac{1}{4\cdot6}+\frac{1}{6\cdot8}+...+\frac{1}{48\cdot50}\)
\(A=\frac{1}{2}\left[\frac{2}{2\cdot4}+\frac{2}{4\cdot6}+\frac{2}{6\cdot8}+...+\frac{2}{48\cdot50}\right]\)
\(A=\frac{1}{2}\left[\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{48}-\frac{1}{50}\right]\)
\(A=\frac{1}{2}\left[\frac{1}{2}-\frac{1}{50}\right]=\frac{1}{2}\cdot\frac{12}{25}=\frac{12}{50}\)
Tương tự bài trên
B = \(\frac{1}{16}+\frac{1}{48}+\frac{1}{96}+...+\frac{1}{880}\)
= \(2.\left(\frac{1}{8}+\frac{1}{24}+\frac{1}{48}+...+\frac{1}{440}\right)\)
= \(2.\left(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{20.22}\right)\)
= \(2.\frac{1}{2}.\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{20.22}\right)\)
= \(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{20}-\frac{1}{22}\)
= \(\frac{1}{2}-\frac{1}{22}\)
= \(\frac{5}{11}\)
\(A=\frac{1}{2.4}+\frac{1}{4.6}+....+\frac{1}{48.50}\)
\(A=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+....+\frac{1}{48}-\frac{1}{50}\)
\(A=\frac{1}{2}-\frac{1}{50}\)
\(A=\frac{25}{50}+\frac{\left(-1\right)}{50}\)
\(A=\frac{24}{50}\)
\(A=\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{48.50}\)
\(A=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{48}-\frac{1}{50}\)
\(A=\frac{1}{2}-\frac{1}{50}\)
\(A=\frac{1}{2}+\frac{-1}{50}=\frac{25}{50}+\frac{-1}{50}=\frac{25+\left(-1\right)}{50}\)
\(A=\frac{24}{50}\)
