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Thread: recurrence relation

  1. #1
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    recurrence relation

    Can anyone tell me if this is correct for this relation:

    6.1 In the following sequences determine s5 if s0, s1, ... sn, ... is a sequence satisfying the given recurrence relation and initial condition.
    b. sn = -sn-1 - n2 for n >= 1, s0 = 2
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  2. #2
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    Sorry forgot to put in the answers. Here is the answers I got:

    s0 = 2
    s1 = -1
    s2 = 5
    s3 = 4
    s4 = 12
    s5 = 13
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  3. #3
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    Quote Originally Posted by papa_chango123 View Post
    Sorry forgot to put in the answers. Here is the answers I got:

    s0 = 2
    s1 = -1
    s2 = 5
    s3 = 4
    s4 = 12
    s5 = 13
    No,
    s1=-s0-1^2=-2-1=-3
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  4. #4
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    Hello, papa_chango123!

    What is $\displaystyle n2$?
    Is that your way of writing $\displaystyle n^2$ ?


    6.1) In the following sequences determine $\displaystyle s_5$
    if $\displaystyle s_n$ is a sequence satisfying the given recurrence relation and initial condition.
    $\displaystyle b)\;s_n \:= \:-s_{n-1}- n^2$ . for $\displaystyle n \geq 1,\;s_o = 2$

    Am I reading it correctly?

    It says: .$\displaystyle \underbrace{s_n}_{\text{the nth term}} \;\underbrace{=}_{\text{is}}\;\underbrace{-s_{n-1}}_{\text{neg.of preceding term}} - \underbrace{n^2}_{\text{minus n-squared}}$

    So $\displaystyle s_1$ is the negative of $\displaystyle s_o$, minus $\displaystyle 1^2.$
    . . $\displaystyle s_1\:=\:-2 - 1^2\:=\:-3$

    And $\displaystyle s_2$ is the negative of $\displaystyle s_1$, minus $\displaystyle 2^2.$
    . . $\displaystyle s_2\:=\:-(-3) - 2^2 \:=\:-1$

    Then $\displaystyle s_3$ is the negative of $\displaystyle s_2$, minus $\displaystyle 3^2.$
    . . $\displaystyle s_3\:=\:-(-1) - 3^2 \:=\:-8$

    Hence $\displaystyle s_4$ is the negative of $\displaystyle s_3$. minus $\displaystyle 4^2.$
    . . $\displaystyle s_4\:=\:-(-8) - 4^2\:=\:-8$

    Therefore: $\displaystyle s_5$ is the negative of $\displaystyle s_4$, minus $\displaystyle 5^2.$
    . . $\displaystyle s_5\:=\:-(-8) - 5^2\:=\:\boxed{-17}$

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  5. #5
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    Yes thanks that is the answer I got.
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