Question 1209829: Find the ordered pair (p,q) such that
F_n = p \alpha^n + q \beta^n.
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Let $F_n$ denote the $n$-th Fibonacci number, where $F_0 = 0$, $F_1 = 1$, and $F_{n+2} = F_{n+1} + F_n$ for $n \ge 0$. The characteristic equation for this recurrence is $x^2 - x - 1 = 0$. The roots of this equation are
\[ \alpha = \frac{1 + \sqrt{5}}{2} \quad \text{and} \quad \beta = \frac{1 - \sqrt{5}}{2}. \]
We are given that $F_n = p \alpha^n + q \beta^n$. We need to find the values of $p$ and $q$.
We can use the initial values $F_0 = 0$ and $F_1 = 1$.
For $n=0$, we have $F_0 = p \alpha^0 + q \beta^0 = p + q = 0$.
For $n=1$, we have $F_1 = p \alpha^1 + q \beta^1 = p \alpha + q \beta = 1$.
From the first equation, $q = -p$. Substituting this into the second equation, we get
\[ p \alpha - p \beta = 1 \]
\[ p (\alpha - \beta) = 1 \]
\[ p \left( \frac{1 + \sqrt{5}}{2} - \frac{1 - \sqrt{5}}{2} \right) = 1 \]
\[ p \left( \frac{2 \sqrt{5}}{2} \right) = 1 \]
\[ p \sqrt{5} = 1 \]
\[ p = \frac{1}{\sqrt{5}}. \]
Since $q = -p$, we have $q = -\frac{1}{\sqrt{5}}$.
Therefore, the ordered pair $(p,q)$ is
\[ \left( \frac{1}{\sqrt{5}}, -\frac{1}{\sqrt{5}} \right). \]
Thus, the formula for the $n$-th Fibonacci number is
\[ F_n = \frac{1}{\sqrt{5}} \left( \frac{1 + \sqrt{5}}{2} \right)^n - \frac{1}{\sqrt{5}} \left( \frac{1 - \sqrt{5}}{2} \right)^n. \]
Final Answer: The final answer is $\boxed{\left( \frac{1}{\sqrt{5}}, -\frac{1}{\sqrt{5}} \right)}$
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