5. Let the sequence $a_{1}, a_{2}, a_{3} \cdots a_{n}, \cdots$ satisfy $a_{1}=a_{2}=1, a_{3}=2$, and for any natural number $n$, $a_{n} \cdot a_{n+1} \cdot a_{n+2} \neq$ 1, and also $a_{n} \cdot a_{n+1} \cdot a_{n+2} \cdot a_{n+3}=a_{n}+a_{n+1}+a_{n+2}+a_{n+3}$, then the value of $a_{1}+a_{2}+\cdots+a_{100}$ is $\qquad$
5. 200 Because $a_{n} a_{n+1} a_{n+2} a_{n+3}=a_{n}+a_{n+1}+a_{n+2}+a_{n+3}$

So $a_{n+1} a_{n+2} a_{n+3} a_{n+4}=a_{n+1}+a_{n+2}+a_{n+3}+a_{n+4}$
Subtracting the two equations gives: $\left(a_{n}-a_{n+4}\right)\left(a_{n+1} a_{n+2} a_{n+3}-1\right)=0$, since $a_{n+1} a_{n+2} a_{n+3} \neq 1$.
Therefore, $a_{n}=a_{n+4}$, which means $\left\{a_{n}\right\}$ is a periodic sequence with a period of 4.
Given $a_{1}=a_{2}=1, a_{3}=2$, from (1) $2 a_{4}=4+a_{4}$
So $a_{4}=4$, thus $a_{1}+a_{2}+a_{3}+a_{4}=8$, so $a_{1}+a_{2}+\cdots+a_{100}=25\left(a_{1}+a_{2}+a_{3}+a_{4}\right)=25 \times$ $8=200$.