00:01
So after the nth stage, the total amount of the table that has been removed is equal to l multiplied by one fourth plus one eighth plus one -sixth all the way up to 1 over 2 to the power of n plus 1.
00:19
So at the first stage at n is equal to 1, there are sections of each length l minus l over 4 divided by 2, which is equal to, let's see, which is equal to 3l divided by 8, which is less than l over 2.
00:37
Suppose that at the k stage, each of the 2 to the power of k remaining sections has length that is less than l over 2 to the power of k.
00:46
Thus, at the k plus 1 stage is obtained by removing the section of length l over 4 to the power of k plus 1, centered at the midpoint of each segment in the k stage.
00:58
So if we let a sub k and a subk plus 1, such as the length of each segment in the k and k plus 1 stage, then we can write 8 subk plus 1 is equal to a sub k minus l over 4 to the power of k plus 1 divided by 2, which is less than l divided by 2 to the power of k minus l over 4 to the power of k plus 1, all divided by 2.
01:28
Therefore, we can write that a sub k, a subk, is less than l over 2 to the power of k, multiplied by 1 minus 1 over 2 to the power of k plus 2, all divided by 2, which is less than l over 2 to the power of k, multiplied by 1 half, which is equal to l over 2 to the power of k plus 1...