Question

Prove the following statements with either induction, strong induction or proof by smallest counterexample. The indicated diagonals of Pascal's triangle sum to Fibonacci numbers. Prove that this pattern continues forever.

Short answer

The proof bases on showing that base cases, n = 0 and 1, are valid, and then using the inductive step to prove that the conjecture remains valid for all further steps. The n-th Fibonacci number equals the sum of the numbers of the n-th row in Pascal's triangle. Therefore, the sum of the numbers in the (k + 1)th diagonal of Pascal's triangle equals the (k + 1)st Fibonacci number. Hence, the statement is true for all integers.

Step-by-step solution

Basics of the problem

First, let's quickly review the Fibonacci sequence and Pascal's triangle. The Fibonacci sequence is a series of numbers in which each number is the sum of the two preceding ones, usually starting with 0 and 1. That is, 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, and so on. Pascal's triangle is a triangular array of the binomial coefficients.

Proof - Base Case

In case of the Fibonacci sequence, the base case are simple. When n = 0, the Fibonacci number is 0. When n = 1, the Fibonacci number is 1. For Pascal's triangle, when n = 0 and 1, the diagonals also sum to 0 and 1, respectively. So, the statement holds for these first two Fibonacci numbers.

Proof - Inductive Step

Assume the statement is true for n = k >= 1. This means that the kth Fibonacci number equals the sum of the numbers on the kth diagonal of Pascal's triangle. Now, we need to prove that the statement is also true for n = k + 1. For Pascal's triangle, each element of the triangle can be represented as \(\binom{n}{r}\) where n is the row number (starting from 0) and r is the term number. Using this, we can prove that the sum of elements of the (k + 1)th diagonal is equal to the (k + 1)st Fibonacci number.

Proof - Conclusion

By principle of mathematical induction, if the statement is true for n = 0 and 1 (base case) and if the statement true for n = k implies that the statement is also true for n = k + 1 (inductive step), then the statement is true for all positive integers. So, we have proven that the indicated diagonals of Pascal's triangle always sum to Fibonacci numbers indefinitely.