[tex]\dfrac{1}{1\cdot 2\cdot 3}+\dfrac{1}{2\cdot 3\cdot 4}+...+\dfrac{1}{2011\cdot 2012\cdot 2013}=\\ \\ = \sum\limits_{n=1}^{2011} \dfrac{1}{n(n+1)(n+2)} =\sum\limits_{n=1}^{2011} \dfrac{1}{n(n+2)}\cdot \dfrac{1}{n+1}} = \\ \\ = \dfrac{1}{2}\sum\limits_{n=1}^{2011} \Big(\dfrac{1}{n}-\dfrac{1}{n+2}\Big)\cdot \dfrac{1}{n+1} = \dfrac{1}{2}\sum\limits_{n=1}^{2011} \Big(\dfrac{1}{n}\cdot \dfrac{1}{n+1}- \dfrac{1}{n+1}\cdot \dfrac{1}{n+2}\Big) =[/tex]
[tex]=\dfrac{1}{2}\sum\limits_{n=1}^{2011} \Big(\dfrac{1}{n(n+1)}- \dfrac{1}{(n+1)(n+2)}}\Big)= \\ \\ = \dfrac{1}{2}\Big(\dfrac{1}{1\cdot 2}+\dfrac{1}{2\cdot 3}+...+\dfrac{1}{2011\cdot 2012}-\dfrac{1}{2\cdot 3}-\dfrac{1}{3\cdot 4}-...-\dfrac{1}{2012\cdot 2013}\Big) = \\ \\ = \dfrac{1}{2}\Big(\dfrac{1}{1\cdot 2}-\dfrac{1}{2012\cdot 2013}}\Big) = \dfrac{4024\cdot 2013-4}{4\cdot 2013\cdot 4024} = \dfrac{8100308}{32401248}[/tex]
