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Math Help - Express Sn as a function of n.

  1. #1
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    Express Sn as a function of n.

    I need to figure out this function before I can do the actual problem that concerns the function. I don't know why I'm having so much trouble with this, but I just can't figure it out!

    S(0) = 1, S(1) = 0, and S(n+1) = 3S(n) 2S(n-1)
    Express S(n) as a function of n.

    And the letters/numbers in parenthesis are supposed to be subscripts of each S.

    These are the first 6 numbers in the sequence.
    S(0) = 1
    S(1) = 0
    S(2) = -2
    S(3) = -6
    S(4) = -14
    S(5) = -30
    S(6) = -62

    Any help?
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  2. #2
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    Re: Express Sn as a function of n.

    Quote Originally Posted by jessaray View Post
    I need to figure out this function before I can do the actual problem that concerns the function. I don't know why I'm having so much trouble with this, but I just can't figure it out!

    S(0) = 1, S(1) = 0, and S(n+1) = 3S(n) 2S(n-1)
    Express S(n) as a function of n.

    And the letters/numbers in parenthesis are supposed to be subscripts of each S.

    These are the first 6 numbers in the sequence.
    S(0) = 1
    S(1) = 0
    S(2) = -2
    S(3) = -6
    S(4) = -14
    S(5) = -30
    S(6) = -62

    Any help?
    What have you tried? As always there is a bit of trial and error when looking for a pattern. Starting with S_2 what is the difference between the sucessive terms? Look and see if you can see a pattern.
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  3. #3
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    Prove It's Avatar
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    Re: Express Sn as a function of n.

    Quote Originally Posted by jessaray View Post
    I need to figure out this function before I can do the actual problem that concerns the function. I don't know why I'm having so much trouble with this, but I just can't figure it out!

    S(0) = 1, S(1) = 0, and S(n+1) = 3S(n) 2S(n-1)
    Express S(n) as a function of n.

    And the letters/numbers in parenthesis are supposed to be subscripts of each S.

    These are the first 6 numbers in the sequence.
    S(0) = 1
    S(1) = 0
    S(2) = -2
    S(3) = -6
    S(4) = -14
    S(5) = -30
    S(6) = -62

    Any help?
    You have \displaystyle S_0 = 1, S_1 = 0, S_{n+1} = 3S_n - 2S_{n-1}

    Assume there's a solution of the form \displaystyle S_n = r^n, then substituting into the recurrence relationship will give

    \displaystyle \begin{align*} r^{n+1} &= 3r^n - 2r^{n-1} \\ r^{n-1}r^2 &= 3r^{n-1}r - 2r^{n-1} \\ r^2 &= 3r - 2 \\ r^2 - 3r + 2 &= 0 \\ (r - 1)(r - 2) &= 0 \\ r = 1\textrm{ or }r &= 2 \end{align*}

    So the solution is

    \displaystyle \begin{align*} S_n &= C_11^n + C_22^n \\ S_n &= C_1 + C_2 2^n \end{align*}

    and now using the initial conditions \displaystyle S_0 = 1 and \displaystyle S_1 = 0 , we find

    \displaystyle 1 = C_1 + C_2 2^0 \implies 1 = C_1 + C_2 and \displaystyle 0 = C_1 + C_22^1 \implies 0 = C_1 + 2C_2 .

    Solving these equations simultaneously gives

    \displaystyle C_1 = 2, C_2 = -1

    and therefore the solution is

    \displaystyle S_n = 2 - 2^n
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  4. #4
    Behold, the power of SARDINES!
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    Re: Express Sn as a function of n.

    Quote Originally Posted by TheEmptySet View Post
    What have you tried? As always there is a bit of trial and error when looking for a pattern. Starting with S_2 what is the difference between the sucessive terms? Look and see if you can see a pattern.
    Just for completeness I will finish what I had hinted at.
    There is a pattern here. the difference between

    S_3-S_2=4=2^2
    S_4-S_3=8=2^3
    S_5-S_4=16=2^4
    S_6-S_5=32=2^5

    So know we know the solution is of the form mentioned in Prove It's post.
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