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Alternate Fibonacci Ratio

Problem

The Fibonacci sequence is defined by the second order recurrence relation $F_{n+2} = F_{n+1} + F_n$, where $F_1 = 1$ and $F_2 = 1$.

$1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, ...$

Consider the ratio of adjacent terms, $\dfrac{F_{n+1}}{F_{n}}$:

1/1 = 1
2/1 = 2
3/2 = 1.5
5/3 = 1.666...
8/5 = 1.6
13/8 = 1.625
21/13 = 1.615...
34/21 = 1.619...
55/34 = 1.617...
89/55 = 1.618...

As $n$ increases it is well known that the ratio of adjacent terms $\dfrac{F_{n+1}}{F_{n}}$ tends towards $\phi = \dfrac{\sqrt{5} + 1}{2} \approx 1.618$ (see Fibonacci Ratio).

What does the ratio $\dfrac{F_{n+2}}{F_n}$ tend towards as $n$ increases?


Solution

By definition, $F_{n+2} = F_{n+1} + F_n$.

Therefore $R = \dfrac{F_{n+2}}{F_n} = \dfrac{F_{n+1} + F_n}{F_n} = \dfrac{F_{n+1}}{F_n} + 1$.

Hence as $n \rightarrow \infty$, $R \rightarrow \phi + 1 = \dfrac{\sqrt{5} + 1}{2} + 1 = \dfrac{\sqrt{5} + 3}{2}$.

Show that $\dfrac{F_{n+3}}{F_n} \rightarrow 2 \phi + 1$ and $\dfrac{F_{n+4}}{F_n} \rightarrow 3 \phi + 2$.
Prove in general that $\dfrac{F_{n+k}}{F_n} \rightarrow F_k \phi + F_{k-1}$.

Problem ID: 310 (15 Feb 2007)     Difficulty: 2 Star

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