\(\frac{4}{3.6}+\frac{4}{6.9}+...+\frac{4}{12.15}\)
\(=\frac{4\left(\frac{3}{3.6}+\frac{3}{6.9}+...+\frac{3}{12.15}\right)}{3}\)
\(=\frac{4\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+...+\frac{1}{12}-\frac{1}{15}\right)}{3}\)
\(=\frac{4\left(\frac{1}{3}-\frac{1}{15}\right)}{3}\)
\(=\frac{\frac{16}{15}}{3}=\frac{48}{15}\)