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Math Help - Need Help Fiinding Recursive formula

  1. #1
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    Need Help Fiinding Recursive formula

    I need help finding a definate formula for the sequence described below.

    a_1=2

    a_2=5

    a_(n+2) = 3a_(n+1) - 2a_n

    It is very similar to the Fibonacci sequence.
    Here is as far as I got.

    x^n = (x^2 - 3x +2) = 0

    x = 2, 1

    Now I need to use these numbers to come up with a formula.
    Last edited by nish915; September 24th 2012 at 05:47 PM.
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  2. #2
    MHF Contributor MarkFL's Avatar
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    Re: Need Help Fiinding Recursive formula

    Since the characteristic roots are 1 and 2, you can expect the closed form to be:

    x_n=k_1\cdot1^n+k_2\cdot2^n=k_1+k_2\cdot2^n

    then use your known (initial) values to determine the coefficients k_i.
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  3. #3
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    Re: Need Help Fiinding Recursive formula

    Sorry, but I just can't seem to understand this any more than what I have already done. What is the final formula that I should come up with? Maybe if I know the formula, I can understand how to get there. Where did "k" come from?
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  4. #4
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    Re: Need Help Fiinding Recursive formula

    Quote Originally Posted by nish915 View Post
    Where did "k" come from?
    As Mark said,
    Quote Originally Posted by MarkFL2 View Post
    use your known (initial) values to determine the coefficients k_i.
    k_i are some constants, yet undetermined. You can substitute n = 1 and n = 2 into the equation in post #2 (where x_n should be replaced with a_n) to get two linear equations in k_1 and k_2. For more information on linear recurrence relation see here and here.
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  5. #5
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    Re: Need Help Fiinding Recursive formula

    Thank you both for the help. I just figured it out. The final equation is (3/2)(2^n) - 1.
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  6. #6
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    Re: Need Help Fiinding Recursive formula

    Hello, nish915!

    \text{given: }\:a(1)\:=\:2,\;\;a(2)\:=\:5 ,\;\;a(n+2) \:=\: 3a(n+1) - 2a(n)

    \text{Find a closed form for }a(n).

    We have: . a(n+2) - 3a(n+1) + 2a(n) \:=\:0

    Let X^n = a(n)\!:\;X^{n+2} - 3X^{n+1} + 2X^n \:=\:0

    Divide by X^n\!:\;\;X^2 - 3X + 2 \:=\:0 \quad\Rightarrow\quad (X - 1)(X - 2) \:=\:0 \quad\Rightarrow\quad X \:=\:1,2

    The function is of the form: . a(n) \:=\:A(1^n) + B(2^n) \:=\:A + 2^nB


    Then we have: . \begin{array}{cccccccccc} a(1) = 2: & A + 2B &=& 2 & [1] \\ a(2) = 5: & A + 4B &=& 5 & [2] \end{array}

    Subtract [2]-[1]: . 2B \,=\,3 \quad\Rightarrow\quad B \,=\,\tfrac{3}{2}

    Substitute into [1]: . A + 2(\tfrac{3}{2}) \:=\:2 \quad\Rightarrow\quad A \:=\:\text{-}1


    Therefore: . a(n) \;=\;\text{-}1 + \tfrac{3}{2}\!\cdot\!2^n \;=\;3\!\cdot\!2^{n-1} - 1
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  7. #7
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    Re: Need Help Fiinding Recursive formula

    Quote Originally Posted by nish915 View Post
    Thank you both for the help. I just figured it out. The final equation is (3/2)(2^n) - 1.
    Yes, and as Soroban did, I would choose to write:

    a_n=3\cdot2^{n-1}-1

    to get integral coefficients. It just looks cleaner.
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