\(S=\frac{1}{9}+\frac{1}{45}+\frac{1}{105}+\frac{1}{189}+\frac{1}{297}=\frac{1}{3}\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}\right)\)
\(=\frac{1}{3}.\frac{1}{2}\left(\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+\frac{2}{99}\right)=\frac{1}{6}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\right)\)
\(=\frac{1}{6}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)=\frac{1}{6}\left(1-\frac{1}{11}\right)=\frac{1}{6}.\frac{10}{11}\)
\(=\frac{5}{33}\)
Bài làm:
\(S=\frac{1}{9}+\frac{1}{45}+\frac{1}{105}+\frac{1}{189}+\frac{1}{297}\)
\(S=\frac{1}{6}\left(\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+\frac{2}{99}\right)\)
\(S=\frac{1}{6}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\right)\)
\(S=\frac{1}{6}\left(\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+\frac{9-7}{7.9}+\frac{11-9}{9.11}\right)\)
\(S=\frac{1}{6}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)\)
\(S=\frac{1}{6}\left(1-\frac{1}{11}\right)\)
\(S=\frac{1}{6}.\frac{10}{11}=\frac{5}{33}\)
Vậy \(S=\frac{5}{33}\)
Xin lỗi bạn Xyz nhé, mk ko có chép bài bạn đâu! với lại mk cx ko k sai bài bn đâu nhé!
