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$$\sum_{n=1}^{2012}\sqrt{1+\frac{1}{n^2}+\frac{1}{(n+1)^2}}$$

$$=\sum_{n=1}^{2012}\frac{\sqrt{n^2(n+1)^2+n^2+(n+1)^2}}{n(n+1)}$$

$$=\sum_{n=1}^{2012}\frac{\sqrt{n^2(n+1)^2+2n^2+2n+1}}{n(n+1)}$$

$$=\sum_{n=1}^{2012}\frac{\sqrt{n^2(n+1)^2+2n(n+1)+1}}{n(n+1)}$$

$$=\sum_{n=1}^{2012}\frac{\sqrt{(n(n+1)+1)^2}}{n(n+1)}$$

$$=\sum_{n=1}^{2012}\frac{n(n+1)+1}{n(n+1)}$$

$$=\sum_{n=1}^{2012} 1+\sum_{n=1}^{2012}\frac{1}{n(n+1)}$$

$$=\sum_{n=1}^{2012} 1+\sum_{n=1}^{2012}(\frac{1}{n}-\frac{1}{n+1})$$

$$=2012+(1-\frac{1}{2013})=\frac{4052168}{2013}.\blacksquare$$