Fibonacci numbers and determinant
Fibonacci sequence is a sequence as
\[1, 2, 3, 5, 8, 13, 21, 34, \cdots,\]
it satisfies: \(F_n = F_{n - 1} + F_{n - 2}(n \ge 3), F_1 = 1, F_2 = 2\)
Proof the Fibonacci numbers \(F_n\) can be get from the determinant \[ F_n = {\begin{vmatrix} 1 & -1 & 0 & 0 & \cdots & 0 & 0 & 0\\ 1 & 1 & -1 & 0 & \cdots & 0 & 0 & 0\\ 0 & 1 & 1 & -1 & \cdots & 0 & 0 & 0\\ \vdots & \vdots & \vdots & \vdots & & \vdots & \vdots & \vdots\\ 0 & 0 & 0 & 0 & \cdots & 1 & 1 & -1\\ 0 & 0 & 0 & 0 & \cdots & 0 & 1 & 1\\ \end{vmatrix} }; \]
Use induction can proof it.
Show the formula of Fibonacci numbers
Because \(F_n = F_{n - 1} + F_{n - 2}(n \ge 3), F_1 = 1, F_2 = 2\), suppose \[F_n - aF_{n - 1} = b(F_{n - 1} - aF_{n - 2})\] so \[F_n = (a + b)F_{n - 1} - abF_{n - 2}\] \[F_n - bF_{n - 1} = a(F_{n - 1} - bF_{n - 2})\] then we get \(a + b = 1, ab = -1\), so \[a = \frac{1 + \sqrt{5}}{2}, b = \frac{1 - \sqrt{5}}{2}\] or \[a = \frac{1 - \sqrt{5}}{2}, b = \frac{1 + \sqrt{5}}{2}\] we consider the first case, because \(F_2 = 2, F_1 = 1\), so \[F_n - aF_{n - 1} = b^{n - 2}(2 - a) = b^{n - 2}(1 + b) = b^{n - 2} + b^{n - 1}\] \[F_n - bF_{n - 1} = a^{n - 2}(2 - b) = a^{n - 2}(1 + a) = a^{n - 2} + a^{n - 1}\] so \[F_n = \frac{a^{n} + a^{n - 1} - b^{n} - b^{n - 1}}{a - b}\] because \(ab = -1\), we can get a more elegant conclusion \[F_n = \frac{a^{n + 1} - b^{n + 1}}{a - b}\]