\(\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+...+10}\)
=\(\frac{1}{3}+\frac{1}{6}+...+\frac{1}{55}\)
=\(2\cdot\left(\frac{1}{6}+\frac{1}{12}+...+\frac{1}{110}\right)\)
=\(2\cdot\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}\right)=2\cdot\left(\frac{1}{2}-\frac{1}{11}\right)=\frac{9}{11}\)