Example 7 (1992 National High School League Question) Let the sequence $a_{1}, a_{2}, \cdots, a_{n}, \cdots$ satisfy $a_{1}=a_{2}=1, a_{3}=2$, and for any positive integer $n$, $a_{n} a_{n+1} a_{n+2} \neq 1$. Also, $a_{n} a_{n+1} a_{n+2} a_{n+3}=a_{n}+a_{n-1}+a_{n+2}+a_{n+3}$, then the value of $a_{1}+a_{2}+\cdots+a_{100}$ is $\qquad$.
Solve: From $a_{n} a_{n+1} a_{n+2} a_{n+3}=a_{n}+a_{n-1}+a_{n+2}+a_{n+3}$, we get $a_{n+1} a_{n+2} a_{n+3} a_{n+4}=a_{n+1}+a_{n+2}+a_{n+3}+a_{n+1}$. Subtracting the two equations, we have $a_{n-1} a_{n+2} a_{n+3}(a_{n}-a_{n+1})=a_{n}-a_{n+1}$. Therefore, $(a_{n}-a_{n+4})(a_{n+1} a_{n+2} a_{n+3}-1)=0$. Since $a_{n+1} a_{n+2} a_{n+3} \neq 1$, it follows that $a_{n}=a_{n+4}$. Hence, $\{a_{n}\}$ is a periodic sequence with a period of 4. Given $a_{1}=a_{2}=1, a_{3}=2, a_{1} a_{2} a_{3} a_{1}=a_{1}+a_{2}+a_{3}+a_{4}$, we have $a_{4}=4$. Therefore, $a_{1}+a_{2}+\cdots+a_{100}=25(a_{1}+a_{2}+a_{3}+a_{4})=25 \times 8=200$.

Explanation: This problem involves manipulating the given recurrence relation of the sequence by changing the indices to construct a corresponding equation. By using the coordination of the two equations (adding or subtracting), some terms are eliminated, leading to a relationship between two terms in the sequence, revealing its periodicity and facilitating the solution. The conclusion of this problem can be generalized to a more general case:

Theorem: If the sequence $\{a_{n}\}$ satisfies $\prod_{i=0}^{m} a_{n+i}=P \sum_{i=0}^{m} a_{n+i} (P \neq 0, \prod_{i=1}^{m} a_{n+1} \neq P, m \in \mathbf{N}^{+})$, then $m+1$ is one of its periods.
Proof: From
$$
\begin{array}{l}
\prod_{i=0}^{m} a_{n+i}=P \sum_{i=0}^{m} a_{n+i}, \\
\prod_{i=1}^{m+1} a_{n+i}=P \sum_{i=1}^{m-1} a_{n+i}.
\end{array}
$$
Subtracting (2) from (1), we get
$$
\left(a_{n-m+1}-a_{n}\right) \prod_{i=1}^{m} a_{n+i}=P a_{n, m+1}-P a_{n},
$$
Thus,
$$
\left(a_{n+m+1}-a_{n}\right)\left(\prod_{i=1}^{m} a_{n+1}-P\right)=0.
$$
Since
$$
\prod_{i=1}^{m} a_{n+i}-P \neq 0, \text{ it follows that } a_{n+m+1}=a_{n}.
$$
Therefore, $m+1$ is a period of the sequence.
Note: $m+1$ is a period of the sequence, but it is not necessarily the smallest positive period. For example, if the first $m+1$ terms of the sequence in the theorem itself have periodicity, the smallest positive period of the sequence is not $m+1$.

In high school, the general method for solving periodic sequence problems is to list the first few terms to observe their periodicity and then use the periodicity to solve the problem. Clearly, the method of listing the first few terms can only solve problems with a small smallest positive period. For sequences with a large smallest positive period, this method is not practical, and deriving that a sequence is periodic based on the first few terms lacks scientific proof. The theorem above provides a scientific basis.