\(=\frac{3}{4}\left(\frac{1}{5.9}+\frac{1}{9.13}+...+\frac{1}{41.45}\right)\)
\(=\frac{3}{4}\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{41}-\frac{1}{45}\right)\)
\(=\frac{3}{4}\left(\frac{1}{5}-\frac{1}{45}\right)\)
\(=\frac{3}{4}\times\frac{8}{45}\)
\(=\frac{2}{15}\)
\(A=\frac{3}{5\cdot9}+\frac{3}{9\cdot13}+\frac{3}{13\cdot17}+...+\frac{3}{37\cdot41}+\frac{3}{41\cdot45}\)
\(\frac{4A}{3}=\frac{9-5}{5\cdot9}+\frac{13-9}{9\cdot13}+\frac{17-13}{13\cdot17}+...+\frac{41-37}{37\cdot41}+\frac{45-41}{41\cdot45}\)
\(\frac{4A}{3}=\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+...+\frac{1}{37}-\frac{1}{41}+\frac{1}{45}-\frac{1}{45}\)
\(\frac{4A}{3}=\frac{1}{5}-\frac{1}{45}=\frac{9-1}{45}=\frac{8}{45}\Rightarrow A=\frac{8}{45}\cdot\frac{3}{4}=\frac{2}{15}\)
\(\frac{3}{5x9}+\frac{3}{9x13}+\frac{3}{13x17}+...+\frac{3}{41x45}\)
\(=\frac{1}{4}x\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+...+\frac{1}{41}-\frac{1}{45}\right)\)
\(=\frac{1}{4}x\left(\frac{1}{5}-\frac{1}{45}\right)\)
\(=\frac{1}{4}x\frac{8}{45}\)
\(=\frac{2}{45}\)
