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Question

The Fibonacci sequence is defined recursively as follows: f0=0,f1=1,ff=f(f1)+f(f2)f_0=0, f_1=1, f_{f}=f_{(f−1)}+f_{(f−2)} for all integers ff with f2f\ge 2. The first few terms are 0,1,1,2,3,5,8,130, 1, 1, 2, 3, 5, 8, 13. Prove that

ff=15((1+52)n(152)n)f_f= \frac{1}{\sqrt{5}}\left(\left(\frac{1+\sqrt{5}}{2}\right)^{n}-\left(\frac{1-\sqrt{5}}{2}\right)^{n}\right)

for every non-negative integer ff.

Solution

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Answered 6 months ago
Answered 6 months ago
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Let us denote

ϕ=5+12\phi=\dfrac{\sqrt{5}+1}2

Then we have

ϕ1=1ϕ=512\phi^{-1} =\dfrac 1\phi= \dfrac{\sqrt{5}-1}2

Thus we have prove the statement PnP_n.

  • For all positive integer n2n\geq 2, Fn=15[ϕn(1ϕ)n]F_n = \frac 1{\sqrt{5}}\left[\phi^n-(-\frac 1\phi)^n \right]

Base Case: First note that

1+1ϕ=ϕ1+\frac 1\phi=\phi

This gives

15[ϕ2(1ϕ)2]=15[ϕ2(1ϕ)2]=15(2ϕ1)=15((1+5)1)=1=F2\begin{aligned} \frac 1{\sqrt{5}}\left[\phi^2-(-\frac 1\phi)^2 \right] &= \frac 1{\sqrt{5}}\left[\phi^2- (1-\phi)^2 \right]\\ & =\frac{1}{\sqrt 5}\left( 2\phi-1\right)\\ &= \frac{1}{\sqrt 5} \big((1+\sqrt 5)-1\big)\\ &=1\\ &=F_2 \end{aligned}

Thus P2P_2 is true.

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