Mathematical induction, hehe, I haven't used this thing for a long time. I'm a little interested in this topic today, and the specific answer is as follows (don't scold me for not doing it for a long time, I forgot the format):

Solution: (1) Title Description: Any two circles are at two intersections, and any three circles are not at the same point.

f( 1)=2，

f(2)=4，

f(3)=8，

f(4)= 14，

(2) Guess f(n)=n*(n- 1)+2.

It is proved that the conjecture holds when 1, n= 1 and f(n)=2.

2. If n=k, f(k)=k*(k- 1)+2 holds.

Then f(k+ 1) now proves to be true.

When the K+ 1 circle is cut with the original k circles, what is the increased area? This is the key to this problem. When the k+ 1 th circle intersects with k circles, there will be 2(k- 1) extra regions, a K+ 1 public * * region and all regions outside the circle will also be divided, the public * * * region will be divided every time, and the external region will also be divided every time. It means that the extra area should be 2(k- 1)+2=2k, which is hard to understand.

So f(k+ 1)= f(k)+2k.

=K*(k- 1)+2+2k

=k*k+k+2

=k(k+ 1)+2

=(K+ 1)(K+ 1- 1)+2

F(k+ 1) also holds.

That is, when n is a natural number, f(n)=n(n- 1)+2 holds.

Write so hard, add some points.