If $$a⊕b= \frac{1}{ \dfrac{1}{a}+ \dfrac{1}{b}}$$, then what is the value of $$ (1\times2)⊕(2\times3)⊕(3 \times4)⊕\ldots⊕(2013\times2014)$$? 
Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations
Notice that $$x\_{1}⊕x\_{2}⊕\ldots⊕x\_{n}={{(\sum\nolimits\_{i=1}^{n}{x\_{i}^{-1}})}^{-1}}$$,  $$\sum\nolimits\_{i=1}^{n}{\frac{1}{i(i+1)}}=\sum\nolimits\_{i=1}^{n}{(\frac{1}{1}}-\frac{1}{i+1})=1-\frac{1}{n+1}=\frac{1}{n+1}$$ and so  $$(1\times 2)\oplus (2\times 3)\oplus (3\times 4)\oplus \cdots \oplus (2013\times 2014)={{(\frac{2013}{2014})}^{-1}}=\frac{2014}{2013}$$,  This is the reason behind the pattern. 