jmc

With summation notation, $S={\sum }_{i=1}^{2007}\sqrt{1+\frac{1}{i^2}+\frac{1}{(i+1{)}^2}}$. By using a common denominator and simplifying, we have $$\begin{array}{rl}S& =\sum _{i=1}^{2007}\sqrt{\frac{i^2(i^2+2i+1)+i^2+2i+1+i^2}{i^2(i+1{)}^2}}\\ & =\sum _{i=1}^{2007}\sqrt{\frac{i^4+2i^3+3i^2+2i+1}{i^2(i+1{)}^2}}\\ & =\sum _{i=1}^{2007}\sqrt{\frac{(i^2+i+1{)}^2}{i^2(i+1{)}^2}}\\ & =\sum _{i=1}^{2007}\frac{i^2+i+1}{i^2+i}\\ & =\sum _{i=1}^{2007}(1+\frac{1}{i(i+1)})\\ & =\sum _{i=1}^{2007}(1+\frac{1}{i}-\frac{1}{i+1})\end{array}$$ Notice that part of the terms telescope, making calculation simpler. Calculation results in $S=2007+1-\frac{1}{2008}$. Thus, $S^2=(2008-\frac{1}{2008}{)}^2=4032064-2+(\frac{1}{2008}{)}^2$. Since $0<(\frac{1}{2008}{)}^2<1$, we conclude that $\lfloor S^2\rfloor =\boxed{4032062}$.