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Question

# The Fibonacci sequence is defined recursively as follows: $f_0=0, f_1=1, f_{f}=f_{(f−1)}+f_{(f−2)}$ for all integers $f$ with $f\ge 2$. The first few terms are $0, 1, 1, 2, 3, 5, 8, 13$. Prove that$f_f= \frac{1}{\sqrt{5}}\left(\left(\frac{1+\sqrt{5}}{2}\right)^{n}-\left(\frac{1-\sqrt{5}}{2}\right)^{n}\right)$for every non-negative integer $f$.

Solution

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Step 1
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Let us denote

$\phi=\dfrac{\sqrt{5}+1}2$

Then we have

$\phi^{-1} =\dfrac 1\phi= \dfrac{\sqrt{5}-1}2$

Thus we have prove the statement $P_n$.

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

Base Case: First note that

$1+\frac 1\phi=\phi$

This gives

\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 $P_2$ is true.

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