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Math Help - Recurrence Relation

  1. #1
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    Recurrence Relation

    Could anyone direct me on how to solve these? Im stuck. (n-1 are subscripts)

    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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  2. #2
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    Even a general method would help! Thanks
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  3. #3
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    Hello, vexiked!

    I'll walk through the first one.


    Could anyone direct me on how to solve these?

    (1)\;A_n\:=\:2A_{n-1} -3,\quad a_0 = -1

    We conjecture that A_n is of the form: . X^n, an exponential expression.

    \begin{array}{ccccc}\text{Then we have:} & X^n &=& 2X^{n-1} - 3 & {\color{blue}[1]} \\<br />
\text{Write the "next" term:} & X^{n+1} &=& 2X^n - 3 & {\color{blue}[2]}\end{array}

    Subtract [2] - [1]: . X^{n+1} - X^n \:=\:2X^n - 2X^{n-1} \quad\Rightarrow\quad X^{n+1} - 3X^n + 2X^{n-1} \:=\:0

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

    . . Hence, the function has: . 1^n\text{ and }2^n


    Form a linear combination of the two roots: . f(n) \;=\;A\cdot1^n + B\cdot2^n

    . . and we have: . f(n) \;=\;A + B\!\cdot\!2^n


    Use the first two terms of the sequence:

    . . \begin{array}{ccccccccc}f(0) = \text{-}1\!: & A + B\!\cdot\!2^0 &=& \text{-}1 & \Rightarrow & A + B &=& \text{-}1 & {\color{blue}[3]} \\<br />
f(1) = \text{-}5\!: & A + B\!\cdot\!2^1 &=&\text{-}5 & \Rightarrow & A + 2B &=& \text{-}5 & {\color{blue}[4]} \end{array}

    Subtract [4] - [3]: . B \:=\:\text{-}4

    Substitute into [3]: . A - 4 \:=\:\text{-}1 \quad\Rightarrow\quad A \:=\:3


    Therefore: . f(n) \:=\:3 - 4\!\cdot\!2^n \quad\Rightarrow\quad\boxed{ f(n) \:=\:3 - 2^{n+2}}

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