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Math Help - Introducing myself + recurrence problem

  1. #1
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    Introducing myself + recurrence problem

    Hi. I'm Paul from Romania. I joined this forum because i wanted to share my knowledge and learn from you all.

    I have a small problem.
    Is there any particular formula to find the general term of a recurrence relation?
    I must find it for this one:

    a1=1
    a2=5
    an+2=an+1 + 2an
    --------------------
    an=?
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  2. #2
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    Re: Introducing myself + recurrence problem

    See Wikipedia.
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  3. #3
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    Re: Introducing myself + recurrence problem

    Hello, Paul!

    Welcome aboard!
    Here is a rather primitive method for this problem.


    a_1\,=\,1\quad a_2 \,=\,5\quad a_{n+2}\:=\:a_{n+1} + 2a_n

    \text{Find }a_n.

    We have: . a_{n+2} - a_{n+1} - 2a_n \;=\;0

    We conjecture that: . X^n \,=\,a_n
    . . (The general term is exponential in nature.)

    Then we have: . X^{n+2} - X^{n+1} - 2X^n \;=\;0

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

    Hence: . X \:=\:\text{-}1,\,2


    The function seems to be: . f(n) = (\text{-}1)^n\,\text{ or }\,f(n) = 2^n

    Assume that f(n) is a linear combination of these two functions.
    . . That is: . f(n) \;=\;A(\text{-}1)^n + B(2^n)


    We know the first two terms of the sequence:

    . . \begin{array}{cccccccc}f(1) = 1\!: & \text{-}A + 2B &=& 1 & [1] \\ f(2) = 5\!: & A + 4B &=& 5 & [2]\end{array}

    Add [1] and [2]: . 6B \,=\,6 \quad\Rightarrow\quad B \,=\,1

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


    Therefore: . f(n) \;=\;(\text{-}1)^n + 2^n
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