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Math Help - Solutions to recurrence relations

  1. #1
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    Solutions to recurrence relations

    Could anyone point me in the right direction with these two questions, I am lost on where to start.

    An=-An-1, a0 = 5 (Note: n-1 is a subscript)
    An=An-1 + 3, a0 = 1
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  2. #2
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    Hello, vexiked1

    You haven't told us what we're supposed to do.
    But I'll assume we want the formula for the n^{th} term.


    (1)\;\;A_n \:=\:-A_{n-1},\;\; A_0 = 5

    It says: each term is the negative of the preceding term.

    So the sequence is: . 5,\:\text{-}5,\;5,\:\text{-}5,\;5,\:\text{-}5,\;\hdots

    The general term is: . A_n \:=\:(\text{-}1)^{n-1}\cdot 5



    (2)\;\;A_n\:=\:A_{n-1} + 3,\;\;A_0 = 1

    It says: Each term is three more than the preceding term.

    So the sequence is: . 1,\:4,\:7,\:10,\:13,\:\hdots

    The general term is: . A_n \;=\;3n-2

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  3. #3
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    My apologies.... I am to find the solution to the recurrence relations with the given initial conditions. So does this still apply? So the basic approach is to find the sequence first then?

    I am also having trouble with:

    An=2An-1 -3, a0 = -1
    An=(n+1)An-1, a0=2
    An=2nAn-1, a0=3
    An=-An-1 + n - 1, a0 = 7
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