\(=10.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{10.11}\right)\)
\(=10.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+..+\frac{1}{10}-\frac{1}{11}\right)\)
\(=10\left(\frac{1}{2}-\frac{1}{11}\right)=10\cdot\frac{9}{22}=\frac{45}{11}\)