Current location - Training Enrollment Network - Mathematics courses - How many pieces can a watermelon be cut with three knives? How many pieces can three knives cut a watermelon at most? How about four dollars? How about five dollars? How about six dollars? Is there an
How many pieces can a watermelon be cut with three knives? How many pieces can three knives cut a watermelon at most? How about four dollars? How about five dollars? How about six dollars? Is there an
How many pieces can a watermelon be cut with three knives? How many pieces can three knives cut a watermelon at most? How about four dollars? How about five dollars? How about six dollars? Is there any formula? some

2+(n- 1)(n^2+n+6)/6

This is just a guess, which must be supported by mathematical induction. After testing, four planes can only divide the space into 15 parts (three planes can divide eight spaces into two parts, and the fourth plane can't divide all eight planes into two parts), so the answer given on the second floor is wrong.

We know that n straight lines can be divided into 1+n(n+ 1)/2 planes at most. I think it has something to do with the number of n plane subspaces.

N-line plane subspace

1 2 2

2 4 4

3 7 8

4 1 1 15

We see that the number in the second column is equal to the sum of the number above it and the number on its left, while the number in the third column is equal to the sum of the number above it and the number on its left. Based on this, we finally get the result of 2+(n- 1) (n 2+n+6)/6.