jmc

This problem illustrates how algebra can clarify arithmetic. Comparing these quantities directly is a chore. Instead, we note that the first and third choices are both of the form $\frac{n}{n-1}+\frac{n}{n+1}$ for $n=2006$ and $n=2007$. Rewriting this expression algebraically leads to $$\frac{n(n+1)}{n^2-1}+\frac{n(n-1)}{n^2-1}=\frac{2n^2}{n^2-1}=2+\frac{2}{n^2-1}.$$In particular both $A$ and $C$ are larger than 2. It is easy to verify that choice $B$ is equal to 2, so it cannot be the answer. Finally, note that larger values of $n$ produce smaller results, implying that $\boxed{\text{A}}$ is the largest. (It is also possible to guess the answer by trying examples involving much smaller numbers.)