Here’s a quick demonstration of a connection between the Fibonacci sequence and geometric sequences.

The famous Fibonacci sequence starts out 1, 1, 2, 3, 5, 8, 13, … The first two terms are both 1, then each subsequent terms is the sum of the two preceding terms.

A **generalized Fibonacci sequence** can start with any two numbers and then apply the rule that subsequent terms are defined as the sum of their two predecessors. For example, if we start with 3 and 4, we get the sequence 3, 4, 7, 11, 18, 29, …

Let φ be the **golden ratio**, the positive solution to the equation 1 + *x* = *x*^{2}. Let φ’ be the conjugate golden ratio, the negative solution to the same quadratic equation. Then

φ = (1 + √ 5)/2

and

φ’ = (1 − √ 5)/2.

Now let’s look at a generalized Fibonacci sequence starting with 1 and φ. Then our terms are 1, φ, 1 + φ, 1 + 2φ, 2 + 3φ, 3 + 5φ, … Let’s see whether we can simplify this sequence.

Now 1 + φ = φ^{2} because of the quadratic equation φ satisfies. That tells us the third term equals φ^{2}. So our series starts out 1, φ, φ^{2}. This is looking like a geometric sequence. Could the fourth term be φ^{3}? In fact, it is. Since the fourth term is the sum of the second and third terms, it equals φ + φ^{2} = φ(1 + φ) = φ(φ^{2}) = φ^{3}. You can continue this line of reasoning to prove that the generalized Fibonacci sequence starting with 1 and φ is in fact the **geometric sequence **1, φ, φ^{2}, φ^{3}, …

Now start a generalized Fibonacci sequence with φ’. Because φ’ is also a solution to 1 + *x* = *x*^{2}, it follows that the sequence 1, φ’, 1 + φ’, 1 + 2φ’, 2 + 3φ’, … equals the geometric sequence 1, φ’, (φ’)^{2}, (φ’)^{3}, …

## More Fibonacci posts