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Math Help - Generalized Fibonacci proof

  1. #1
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    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,

     <br />
a_n = s \left(\frac{1+\sqrt{5}}{2}\right)^n + t\left(\frac{1-\sqrt{5}}{2}\right)^n<br />

    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

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


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

     <br />
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?

    Thanks a lot!
    Last edited by armeros; April 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 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 \\<br />
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}

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  3. #3
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    Thanks a lot!
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