Example $12 \triangle A B C$ has 1997 points inside it. Using the vertices $A, B, C$ and these 1997 points, how many small triangles can the original triangle be divided into?
Solution: Let the $n-1$ points inside $\triangle A B C$ divide the original triangle into $a_{n-1}$ small triangles. Now consider the situation after adding one more point $P_{n}$:
(1) If point $p_{n}$ is inside a small triangle, then the three vertices of the original small triangle, along with $p_{n}$, divide it into three, i.e., two more small triangles are added.
(2) If point $p_{n}$ is on the common edge of two small triangles, then the vertices of these two small triangles, along with $p_{n}$, divide each of the two small triangles into two, i.e., two more small triangles are added.

Therefore, the number of small triangles into which $n$ points inside triangle $A B C$ divide the original triangle is $a_{n}=a_{n-1}+2$.

Since the sequence $\left\{a_{n}\right\}$ is an arithmetic sequence with the first term $a_{1}=3$ and common difference $d=2$, then $a_{n}=3+(n-1) \times 2=2 n+1$.

Thus, when $n=1997$, the three vertices and these 1997 internal points can divide the original triangle into $2 \times 1997+1=3995$ small triangles.