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Thread: Generalized Fibonacci proof

  1. #1
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    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!
    Last edited by armeros; Apr 23rd 2009 at 09:12 AM.
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  2. #2
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    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: .$\displaystyle a(n) \;=\;\text{-}3\!\cdot\!2^n + 2\!\cdot\!3^n \quad\Rightarrow\quad\boxed{ a(n) \;=\;2\cdot3^n - 3\cdot2^n}$

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  3. #3
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    Thanks a lot!
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