By cross reference # 4137, it spears that the key is to find constant $A$, $B$, and $C$, such that $$\frac{1}{n(n+1)(n+2)}=\frac{A}{n}+\frac{B}{n+1}+\frac{C}{n+2}$$

Rewriting the above relation to the following polynomial by multiplying $n(n+1)(n+2)$: $$1=A(n+1)(n+2) + Bn(n+2) + Cn(n+1)$$

Because this polynomial is expected to be an identity, it must hold for any value of $n$. Letting $n=0$, $-1$, and $-2$, respectively, will lead to $A=\frac{1}{2}$, $B=-1$, and $C=\frac{1}{2}$. We can verify the following relation does hold: $$\frac{1}{n(n+1)(n+2)} = \frac{1}{2}\Big(\frac{1}{n} - \frac{2}{n+1} + \frac{1}{n+2}\Big)$$

Therefore we have $$\begin{array}{rcl} \frac{1}{1\times 2 \times 3} &=& \frac{1}{2}\times\Big(\frac{1}{1} - \frac{2}{2} + \frac{1}{3}\Big)\\ \frac{1}{2\times 3 \times 4} &=& \frac{1}{2}\times\Big(\frac{1}{2} - \frac{2}{3} + \frac{1}{4}\Big)\\ \frac{1}{3\times 4 \times 5} &=& \frac{1}{2}\times\Big(\frac{1}{3} - \frac{2}{4} + \frac{1}{5}\Big)\\ \frac{1}{4\times 5 \times 6} &=& \frac{1}{2}\times\Big(\frac{1}{4} - \frac{2}{5} + \frac{1}{6}\Big)\\ \frac{1}{5\times 6 \times 7} &=& \frac{1}{2}\times\Big(\frac{1}{5} - \frac{2}{6} + \frac{1}{7}\Big)\\ \frac{1}{6\times 7 \times 8} &=& \frac{1}{2}\times\Big(\frac{1}{6} - \frac{2}{7} + \frac{1}{8}\Big)\\ \frac{1}{7\times 8 \times 9} &=& \frac{1}{2}\times\Big(\frac{1}{7} - \frac{2}{8} + \frac{1}{9}\Big)\\ \frac{1}{8\times 9 \times 10} &=& \frac{1}{2}\times\Big(\frac{1}{8} - \frac{2}{9} + \frac{1}{10}\Big) \end{array}$$

When adding these equations, the terms on the right side will cancel along the diagonal direction and only those in the top left and bottom right triangular areas will survive. Therefore, the original expression is equivalent to $$\frac{1}{2}\Big(\frac{1}{1} - \frac{2}{2} + \frac{1}{2} + \frac{1}{9}-\frac{2}{9} + \frac{1}{10}\Big)=\boxed{\frac{11}{45}}$$