5. The numbers $\frac{1}{1}, \frac{1}{2}, \ldots, \frac{1}{2010}$ are written on a blackboard. A student chooses any two of the numbers, say $x, y$, erases them and then writes down $x+y+x y$. He continues to do this until only one number is left on the blackboard. What is this number?
5. We shall prove by induction that if the original numbers are $a_{1}, \ldots, a_{n}, n \geq 2$, then the last number is $\left(1+a_{1}\right) \cdots\left(1+a_{n}\right)-1$.

The assertion is certainly true for $n=2$, the base case. Now suppose it is true for $n=k \geq 2$. Consider $k+1$ numbers $a_{1}, \ldots, a_{k+1}$ written on the board. After one operation, we are left with $k$ numbers. Without loss of generality, we can assume that the student erases $a_{k}$ and $a_{k+1}$ and writes $b_{k}=a_{k}+a_{k+1}+a_{k} a_{k+1}=\left(1+a_{k}\right)\left(1+a_{k+1}\right)-1$. After a further $k$ operations, we are left with the number
$$
\left(1+a_{1}\right) \cdots\left(1+a_{k-1}\right)\left(1+b_{k}\right)-1=\left(1+a_{1}\right) \cdots\left(1+a_{k-1}\right)\left(1+a_{k}\right)\left(1+a_{k+1}\right)-1 .
$$

This completes the proof of the inductive step. Thus the last number is
$$
\left(1+\frac{1}{1}\right)\left(1+\frac{1}{2}\right) \cdots\left(1+\frac{1}{2010}\right)-1=2010
$$