Proving the Product of Squared Fibonacci Numbers

How can we prove that f₁²f₂²...fₙ² = fₙfₙ₋₁...f₁f₀?

Using strong induction, can we prove that the product of the first n Fibonacci numbers squared is equal to the product of the (n+1)th and nth Fibonacci numbers?

Proof Using Strong Induction

Using strong induction, we can prove that the product of the first n Fibonacci numbers squared is equal to the product of the (n+1)th and nth Fibonacci numbers.

Understanding Strong Induction on Fibonacci Numbers

Strong induction is a powerful proof technique that allows us to prove a statement for all positive integers by assuming that it holds true for all previous values up to a specific integer n.

Let's break down the proof of the product of squared Fibonacci numbers. First, we establish the base case for n = 1:

f₁² = f₁ x f₀ = 1 x 1 = f₁f₀

Next, we make the inductive assumption that the statement is true for all values up to n:

f₁²f₂²...fₙ² = fₙfₙ₋₁...f₁f₀

With this assumption in place, we want to show that the statement holds for n+1 as well. By substituting in f(n+1)² for the first term and simplifying using the identity fn+1 = fn + fn-1, we can conclude that the statement is true for n+1 as well.

Therefore, by using strong induction, we have proven that the product of squared Fibonacci numbers is equal to the product of the (n+1)th and nth Fibonacci numbers for all positive integers n.