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Exercise 730Calculus › Single-variable calculus › SequencesL1Added 2 Nov 2026

Binet's formula for the Fibonacci numbers

#Fibonacci#golden ratio#linear recurrence#limits

Statement

Let F0=0F_0=0, F1=1F_1=1 and Fn+2=Fn+1+FnF_{n+2}=F_{n+1}+F_n.

  1. Find a closed formula for FnF_n.
  2. Show that Fn+1Fn→φ=1+52\dfrac{F_{n+1}}{F_n}\to\varphi=\dfrac{1+\sqrt5}2.
Hint

The characteristic equation is r2=r+1r^2=r+1. Write ψ=1−52\psi=\frac{1-\sqrt5}2 for the other root and note that ∣ψ∣<1|\psi|<1.

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Solution

1. The roots of r2−r−1=0r^2-r-1=0 are φ=1+52\varphi=\frac{1+\sqrt5}2 and ψ=1−52\psi=\frac{1-\sqrt5}2. So Fn=Aφn+BψnF_n=A\varphi^n+B\psi^n. From F0=0F_0=0: B=−AB=-A. From F1=1F_1=1: A(φ−ψ)=A5=1A(\varphi-\psi)=A\sqrt5=1. Hence
Fn=φn−ψn5.F_n=\frac{\varphi^n-\psi^n}{\sqrt5}.

2. Since ∣ψ∣<1<φ|\psi|<1<\varphi, we have (ψφ)n→0\left(\frac\psi\varphi\right)^n\to0 and
Dividing numerator and denominator by φn\varphi^n:
Fn+1Fn=φn+1−ψn+1φn−ψn=φ−ψ(ψφ)n1−(ψφ)n⟶φ.\frac{F_{n+1}}{F_n} = \frac{\varphi^{n+1}-\psi^{n+1}}{\varphi^n-\psi^n}=\frac{\varphi-\psi\left(\frac\psi\varphi\right)^n}{1-\left(\frac\psi\varphi\right)^n}\longrightarrow\varphi.