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}