A = \(\frac{1}{15}+\frac{1}{35}+...+\frac{1}{195}+\frac{1}{225}=\)\(\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{13\cdot15}+\frac{1}{225}\)
2A = \(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{13\cdot15}+\frac{2}{225}=\)\(\frac{5-3}{3\cdot5}+\frac{7-5}{5\cdot7}+...+\frac{15-13}{13\cdot15}+\frac{2}{225}\)
2A = \(\frac{5}{3\cdot5}-\frac{3}{3\cdot5}+\frac{7}{5\cdot7}-\frac{5}{5\cdot7}+...+\frac{15}{13\cdot15}-\frac{13}{13\cdot15}+\frac{2}{225}\)
2A = \(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}+\frac{2}{225}\)
2A = \(\frac{1}{3}-\frac{1}{15}+\frac{2}{225}=\frac{60}{225}+\frac{2}{225}=\frac{62}{225}\)
=> A = \(\frac{62}{225}\div2=\frac{31}{225}\)
\(A=\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+...+\frac{1}{195}+\frac{1}{225}\)
\(=\frac{1}{3\times5}+\frac{1}{5\times7}+\frac{1}{7\times9}+...+\frac{1}{13\times15}+\frac{1}{15\times17}\)
\(=\frac{1}{2}\times\left(\frac{2}{3\times5}+\frac{2}{5\times7}+\frac{2}{7\times9}+...+\frac{2}{13\times15}+\frac{2}{15\times17}\right)\)
\(=\frac{1}{2}\times\left(1-\frac{1}{17}\right)=\frac{8}{17}\)
