De rij van Fibonacci

$\begin{array}{l} u_n  = u_{n - 1}  + u_{n - 2} \,\,voor\,\,n \ge 2 \\ met\,\,u_0  = 1\,\,en\,\,u_1  = 1 \\ \end{array}$

Uitgewerkt

$u_n  = u_{n - 1}  + u_{n - 2}$

Neem $u_n  = g^n$:

$\eqalign{  & g^n  = g^{n - 1}  + g^{n - 2}   \cr  & g^2  = g + 1  \cr  & g^2  - g - 1 = 0  \cr  & \left( {g - \frac{1}{2}} \right)^2  - \frac{5}{4} = 0  \cr  & \left( {g - \frac{1}{2}} \right)^2  = \frac{5}{4}  \cr  & g - \frac{1}{2} =  \pm \frac{1}{2}\sqrt 5   \cr  & g = \frac{1}{2} \pm \frac{1}{2}\sqrt 5  \cr}$

Je krijgt dan:

$u_n  = A \cdot \left( {\frac{1}{2} - \frac{1}{2}\sqrt 5 } \right)^n  + B \cdot \left( {\frac{1}{2} + \frac{1}{2}\sqrt 5 } \right)^n$

Vul de startwaarden in:

$\begin{array}{l} \left\{ \begin{array}{l} 0 = A + B \\ 1 = A \cdot \left( {\frac{1}{2} - \frac{1}{2}\sqrt 5 } \right) + B \cdot \left( {\frac{1}{2} + \frac{1}{2}\sqrt 5 } \right) \\ \end{array} \right. \\ \left\{ \begin{array}{l} A = - B \\ 1 = \frac{1}{2}A - \frac{1}{2}A\sqrt 5 + \frac{1}{2}B + \frac{1}{2}B\sqrt 5 \\ \end{array} \right. \\ \left\{ \begin{array}{l} A = - B \\ 2 = A - A\sqrt 5 + B + B\sqrt 5 \\ \end{array} \right. \\ \left\{ \begin{array}{l} A = - B \\ 2 = - B + B\sqrt 5 + B + B\sqrt 5 \\ \end{array} \right. \\ \left\{ \begin{array}{l} A = - B \\ 2 = 2B\sqrt 5 \\ \end{array} \right. \\ \left\{ \begin{array}{l} A = - \frac{1}{{\sqrt 5 }} \\ B = \frac{1}{{\sqrt 5 }} \\ \end{array} \right. \\ \end{array}$

De expliciete formule wordt:

$\eqalign{  & u_n  =  - \frac{1}{{\sqrt 5 }} \cdot \left( {\frac{1}{2} - \frac{1}{2}\sqrt 5 } \right)^n  + \frac{1}{{\sqrt 5 }} \cdot \left( {\frac{1}{2} + \frac{1}{2}\sqrt 5 } \right)^n   \cr   & u_n  = \frac{1}{{\sqrt 5 }} \cdot \left[ {\left( {\frac{1}{2} + \frac{1}{2}\sqrt 5 } \right)^n  + \left( {\frac{1}{2} - \frac{1}{2}\sqrt 5 } \right)^n } \right]  \cr   & u_n  = \frac{{\left( {1 + \sqrt 5 } \right)^n  - \left( {1 - \sqrt 5 } \right)^n }}{{2^n  \cdot \sqrt 5 }} \cr} $