1. Generalized Fibonacci proof

1. A sequence of numbers $\displaystyle a_0, a_1, a_2,...$is defined for every $\displaystyle 2 \leq n, a_n = a_{n-2} + a_{n-1}$.

Prove that there are real numbers $\displaystyle s$and $\displaystyle t$ such that for all natural number $\displaystyle n$,

$\displaystyle a_n = s \left(\frac{1+\sqrt{5}}{2}\right)^n + t\left(\frac{1-\sqrt{5}}{2}\right)^n$

I just plugged in the numbers $\displaystyle s = t = 1$ and it works, as the problem just asked to find some real numbers. But that turns out to be the Lucas squence. In fact, the general answer states that

$\displaystyle s = \frac{(5a_0 +(2a_1 - a_0)\sqrt{5})}{10}$ and $\displaystyle t = \frac{(5a_0 -(2a_1 - a_0)\sqrt{5})}{10}.$
I just have no idea how s and t come.

2. A squence $\displaystyle a_0, a_1, a_2, ...$is defined as follows

$\displaystyle a_0 = -1, a_1 = 0$

for every $\displaystyle 2 \leq n, a_n = 5a_{n-1} - 6a_{n-2}$

Find a formula for $\displaystyle a_n$.

Again, the answer is $\displaystyle a_n = 2 \cdot 3^n - 3 \cdot 2^n$

It looks like I need to apply the formula above, but how?

Thanks a lot!

2. Hello, armeros!

I'll show you the way I handle #2 . . .

2. A sequence $\displaystyle a_0, a_1, a_2, ...$is defined: .$\displaystyle a_0 = \text{-}1,\;a_1 = 0,\;\,a_n \:=\:5a_{n-1} - 6a_{n-2}\text{ for }n \geq 2$

Find a formula for $\displaystyle a_n$.

Answer: .$\displaystyle a_n \:=\:2\!\cdot\!3^n - 3\!\cdot\!2^n$
We conjecture that $\displaystyle a(n)$ is an exponential function: .$\displaystyle a(n) \:=\:X^n$

We have: .$\displaystyle a(n) \:=\:5a_{n-1} - 6a_{n-2} \quad\Rightarrow\quad X^n \:=\:5X^{n-1} - 6X^{n-2} \quad\Rightarrow\quad X^n - 5X^{n-1} + 6X^{n-2} \:=\:0$

Divide by $\displaystyle X^{n-2}\!:\quad X^2 - 5X + 6 \:=\:0 \quad\Rightarrow\quad (X - 2)(X - 3) \:=\:0$

Hence: .$\displaystyle X \:=\:2,\,3$ . . . And we have: .$\displaystyle a(n) \;=\;A\!\cdot\!2^n + B\!\cdot\!3^n$

We know the first values of the sequence:

. . $\displaystyle \begin{array}{ccccccccc}a(0) = \text{-}1\!: & A(2^0) + B(3^0) &=& \text{-}1 & \Rightarrow & A + B &=& \text{-}1 \\ a(1) = 0\!: & A(2^1) + B(3^1) &=& 0 & \Rightarrow & 2A + 3B &=& 0 \end{array}$

Solve the system: .$\displaystyle A \:=\:\text{-}3,\;B \:=\:2$

Therefore: .$\displaystyle a(n) \;=\;\text{-}3\!\cdot\!2^n + 2\!\cdot\!3^n \quad\Rightarrow\quad\boxed{ a(n) \;=\;2\cdot3^n - 3\cdot2^n}$

3. Thanks a lot!