Why is there a Sierpinki's triangle when you shade the multiples of 2 in Pascal's triangle?

Question

How do they know that this pattern will last forever? Why are they so sure that the pattern will not end somewhere?

Answer 1

Consider Pascal's Triangle mod 2:

••••1••••
•••1•1•••
••1•0•1••
•1•1•1•1•
1•0•0•0•1

Sorry about the weird look, but this is hard to do in answers.yahoo.

Now each row adds 1/2 of a place to each side (so 1 place total). Therefore when you continue it is impossible for the 0•0•0 (in the last row) to be filled in in less than 3 rows. Since 0's do not change anything, it is the same as having this triangle:

••••1••••
•••1•1•••
••1•0•1••
•1•1•1•1•

and then starting over at the two bottom corners (where we have the 1's in the original last row). Therefore (after 8 rows) we will get three copies of

••••1••••
•••1•1•••
••1•0•1••
•1•1•1•1•

and in the middle it will be filled with 0's.

Or this:

••••••••1••••••••
•••••••1•1•••••••
••••••1•0•1••••••
•••••1•1•1•1•••••
••••1•0•0•0•1••••
•••1•1•0•0•1•1•••
••1•0•1•0•1•0•1••
•1•1•1•1•1•1•1•1•

Since the bottom will be 1•1•1•1•1•1•1•1, the next row will be 1•0•0•0•0•0•0•0•1. Therefore using the same logic above we will get two more copies of the first 8 rows (with 0's in between). Then we continue and we will get two more copies of the first 16 rows . . . first 32 rows . . . and so on.

Therefore we will have a Sierpinski's triangle.

Answer 2

It will last forever, that is determinsitic