Let $x_{0}, x_{1}, x_{2}, \ldots$ be a sequence where the first two terms are positive, and it holds that $x_{n+2}=\frac{x_{n+1}+1}{x_{n}}$. Express the 2003rd term of the sequence in terms of $x_{0}$ and $x_{1}$.
Solution. If $x_{n}$ and $x_{n+1}$ are positive, then $x_{n+2}$ is meaningful and positive. Since $x_{0}$ and $x_{1}$ are positive, by the principle of mathematical induction, all elements of the sequence will be meaningful and positive numbers.

If $x_{0}=a$ and $x_{1}=b$, then according to the recursive definition, $x_{2}=\frac{b+1}{a}, x_{3}=\frac{a+b+1}{a b}, x_{4}=\frac{a+1}{b}, x_{5}=a, x_{6}=b$.

Thus, $x_{5}=x_{0}, x_{6}=x_{1}$. If for some natural number $k$, $x_{k+5}=x_{k}$ and $x_{k+6}=x_{k+1}$, then $x_{k+7}=\frac{x_{k+6}+1}{x_{k+5}}=$ $\frac{x_{k+1}+1}{x_{k}}=x_{k+2}$. Therefore, the sequence is periodic with a period of 5.

Therefore, the 2003rd term of the sequence is: $x_{2002}=x_{2}=\frac{x_{1}+1}{x_{0}}$.