5. Select an integer sequence $\mathrm{a}_{1}, \mathrm{a}_{2}, \mathrm{a}_{3}, \cdots$, such that:
$$
a_{n}=a_{n-1}-a_{n-2} \text { (where } n \geqslant 3 \text { ) }
$$

If the sum of the first 1492 terms is 1985, and the sum of the first 1935 terms is 1492, what is the sum of the first 2001 terms?
Solve We calculate the first eight terms of this sequence, $a_{1}$,
$$
\begin{array}{l}
a_{2}, a_{3}=a_{2}-a_{1}, a_{4}=a_{3}-a_{2}=a_{2}-a_{1} \\
-a_{2}=-a_{1}, a_{5}=a_{4}-a_{3}=-a_{1}-a_{2}+a_{1} \\
=-a_{2}, a_{6}=a_{5}-a_{4}=a_{1}-a_{2}, a_{7}=a_{0}-a_{5} \\
=\left(a_{1}-a_{2}\right)-\left(-a_{2}\right)=a_{1}, a_{8}=a_{7}-a_{0} \\
=a_{1}-\left(a_{1}-a_{2}\right)=a_{2}, \cdots \text { and we find that every six }
\end{array}
$$

terms form a cycle. For any $n$, we have $a_{n+8}=a_{n}$. It is also easy to prove that in this sequence, the sum of every six consecutive terms is 0. Note that 1492 leaves a remainder of 4 when divided by 6, 1985 leaves a remainder of 5 when divided by 6, and 2001 leaves a remainder of 3 when divided by 6, so
$$
\begin{aligned}
S_{1035} & =a_{1081}+a_{1082}+\cdots+a_{1258} \\
=a_{1}+a_{2} & +\cdots+a_{5} \\
& =a_{1}+a_{2}+\left(a_{2} \cdots a_{1}\right)+\left(-a_{1}\right) \\
& +\left(-a_{1}\right)=a_{2}-a_{1}=1492 . \quad(1)
\end{aligned}
$$
$$
\begin{aligned}
S_{1402} & =a_{140}+a_{1400}+a_{1401}+a_{242} \\
& =a_{1}+a_{2}+a_{3}+a_{4} \\
& =a_{1}+a_{2}+\left(a_{2}-a_{1}\right)+\left(-a_{1}\right) \\
& =2 a_{2}-a_{1}=1985 .
\end{aligned}
$$

From (1) and (2), we solve to get $a_{1}=-999, a_{2}=493$.
$$
\begin{aligned}
S_{2001} & =a_{1090}+a_{2000}+a_{2001} \\
& =a_{1}+a_{2}+a_{3} \\
& =a_{1}+a_{2}+\left(a_{2}-a_{1}\right)=2 a_{2} \\
& =986 .
\end{aligned}
$$