\(\frac{1}{18}+\frac{1}{54}+\frac{1}{108}+...+\frac{1}{990}\)
\(=\frac{1}{3}.\left(\frac{3}{3.6}+\frac{3}{6.9}+\frac{3}{9.12}+...+\frac{3}{30.33}\right)\)
\(=\frac{1}{3}.\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+...+\frac{1}{30}-\frac{1}{33}\right)\)
\(=\frac{1}{3}.\left(\frac{1}{3}-\frac{1}{33}\right)\)
\(=\frac{1}{3}.\frac{10}{33}\)
\(=\frac{10}{99}\)
Đúng không Bạch Dương ?
Ta có: \(\frac{1}{18}+\frac{1}{54}+\frac{1}{108}+...+\frac{1}{990}\)
\(=\frac{1}{2.9}+\frac{1}{6.9}+\frac{1}{12.9}+...+\frac{1}{110.9}\)
\(=\frac{1}{9}\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{110}\right)\)
\(=\frac{1}{9}\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10.11}\right)\)
\(=\frac{1}{9}\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}\right)\)
\(=\frac{1}{9}\left(\frac{1}{1}-\frac{1}{11}\right)\)
\(=\frac{1}{9}.\frac{10}{11}\)
\(=\frac{10}{99}\)
Vậy \(\frac{1}{18}+\frac{1}{54}+\frac{1}{108}+...+\frac{1}{990}=\frac{10}{99}\)
\(=\frac{1}{3x6}+\frac{1}{6x9}+\frac{1}{9x12}+...+\frac{1}{30x33}\)
\(=\frac{1}{3}x\left(\frac{3}{3x6}+\frac{3}{6x9}+....+\frac{3}{30x33}\right)\)
\(=\frac{1}{3}x\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+....+\frac{1}{30}-\frac{1}{33}\right)\)
\(=\frac{1}{3}x\left(\frac{1}{3}-\frac{1}{33}\right)\)
\(=\frac{1}{3}x\frac{10}{33}\)
\(=\frac{10}{99}\)
