# Generalized Fibonacci proof

• Apr 23rd 2009, 09:52 AM
armeros
Generalized Fibonacci proof
1. A sequence of numbers $a_0, a_1, a_2,...$is defined for every $2 \leq n, a_n = a_{n-2} + a_{n-1}$.

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

$
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 $s = t = 1$ 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

$
s = \frac{(5a_0 +(2a_1 - a_0)\sqrt{5})}{10}$
and $t = \frac{(5a_0 -(2a_1 - a_0)\sqrt{5})}{10}.
$

I just have no idea how s and t come. http://www.gconsole.com/forum/emo/5c745924.gif

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

$
a_0 = -1, a_1 = 0$

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

Find a formula for $a_n$.

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

It looks like I need to apply the formula above, but how? http://www.gconsole.com/forum/emo/4519626a.gif

Thanks a lot!
• Apr 23rd 2009, 01:35 PM
Soroban
Hello, armeros!

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

Quote:

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

Find a formula for $a_n$.

Answer: . $a_n \:=\:2\!\cdot\!3^n - 3\!\cdot\!2^n$

We conjecture that $a(n)$ is an exponential function: . $a(n) \:=\:X^n$

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

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

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

We know the first values of the sequence:

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

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

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

• Apr 24th 2009, 08:49 AM
armeros
Thanks a lot!