Question 1209830: Let
A_0 = 0
A_1 = 1
A_n = A_{n - 1} + A_{n - 2} for n ge 2
There is a unique ordered pair (c,d) such that c \alpha^n + d \beta^n is the closed form for sequence A_n. Find the ordered pair (c,d).
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! You've described the Fibonacci sequence, where:
* $A_0 = 0$
* $A_1 = 1$
* $A_n = A_{n-1} + A_{n-2}$ for $n \ge 2$
We're looking for the closed-form expression of $A_n$ in the form $c \alpha^n + d \beta^n$, where $\alpha$ and $\beta$ are the roots of the characteristic equation.
**1. Find the Characteristic Equation and its Roots**
The characteristic equation for the recurrence relation is:
* $x^2 - x - 1 = 0$
Using the quadratic formula, the roots are:
* $\alpha = \frac{1 + \sqrt{5}}{2}$ (the golden ratio)
* $\beta = \frac{1 - \sqrt{5}}{2}$
**2. Use Initial Conditions to Solve for c and d**
We have the following initial conditions:
* $A_0 = 0$
* $A_1 = 1$
Substitute these into the closed-form expression:
* For $n = 0$: $A_0 = c \alpha^0 + d \beta^0 = c + d = 0$
* For $n = 1$: $A_1 = c \alpha^1 + d \beta^1 = c \left( \frac{1 + \sqrt{5}}{2} \right) + d \left( \frac{1 - \sqrt{5}}{2} \right) = 1$
From the first equation, we get $d = -c$. Substitute this into the second equation:
* $c \left( \frac{1 + \sqrt{5}}{2} \right) - c \left( \frac{1 - \sqrt{5}}{2} \right) = 1$
* $c \left( \frac{1 + \sqrt{5} - (1 - \sqrt{5})}{2} \right) = 1$
* $c \left( \frac{2 \sqrt{5}}{2} \right) = 1$
* $c \sqrt{5} = 1$
* $c = \frac{1}{\sqrt{5}}$
Now, substitute $c$ back into $d = -c$:
* $d = -\frac{1}{\sqrt{5}}$
**3. The Ordered Pair (c, d)**
Therefore, the ordered pair $(c, d)$ is:
* $\left( \frac{1}{\sqrt{5}}, -\frac{1}{\sqrt{5}} \right)$
Final Answer: The final answer is $\boxed{\left( \frac{1}{\sqrt{5}}, -\frac{1}{\sqrt{5}} \right)}$
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