CombinatoricsDifficulty 5.1AIME, harderProve itUnited States
Problem:
An infinite table of nonnegative integers is constructed as follows: in the top row, some number is 1 and all other numbers are 0's; in each subsequent row, every number is the sum of some two of the three closest numbers in the preceding row. An example of such a table is shown below.
The top row (with the one 1) is called row 0; the next row is row 1; the next row is row 2, and so forth.
Note that the following problems require you to prove the statements for every table that can be constructed by the process described above, not just for the example shown.
1. Show that any number in row n (for n>0) is at most 2n−1.
Solution
Solution:
We use induction on n. It is clear that any number in row 1 is at most 1=20. Now, if every number in row n is at most 2n−1, then every number in row n+1 is the sum of two numbers in row n and so is at most 2n−1+2n−1=2n. This gives the induction step, and the result follows.
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Source: MathNet,
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