\(A=\frac{7}{9}+\frac{7}{45}+\frac{7}{105}+...+\frac{7}{27645}\)
\(=7\left(\frac{1}{9}+\frac{1}{45}+\frac{1}{105}+...+\frac{1}{27645}\right)\)
\(=7.\frac{1}{3}\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+...+\frac{1}{9215}\right)\)
\(=\frac{7}{3}\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{95.97}\right)\)
Đặt \(S=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{95.97}\)
\(2S=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{95.97}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{95}-\frac{1}{97}\)
\(=1-\frac{1}{97}=\frac{96}{97}\)
\(\Rightarrow S=\frac{96}{97}:2=\frac{96}{97.2}=\frac{48}{97}\). Thay vào \(A\) ta có:
\(A=\frac{7}{3}.\frac{48}{97}=\frac{112}{97}\)
Vậy \(A=\frac{112}{97}\).
