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Math Help - Pattern question is stumping me!

  1. #1
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    Pattern question is stumping me!

    -3, -2, 0, 3

    Can someone tell me what the algebra expression is for this pattern. One variable only. Thanks so much!!!
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  2. #2
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    Quote Originally Posted by squall458 View Post
    -3, -2, 0, 3

    Can someone tell me what the algebra expression is for this pattern. One variable only. Thanks so much!!!

    Hi squall458,

    I think Soroban will come up with an eloquent explanation. I just ran a quadratic regression on my calculator for the sequence:

    Code:
    X:  1    2    3    4    5    6    7
    Y: -3   -2    0    3    7    12   18
    And came up with y=.5x^2-.5x-3
    Last edited by masters; October 20th 2010 at 10:19 AM.
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  3. #3
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    Hello, squall458!

    I'll show you a procedure for this problem.
    It's rather long, but it's simple and direct.


    -3, -2, 0, 3

    Can someone tell me what the algebra expression is for this pattern?

    Take the difference of consecutive terms,
    . . then take difference of the differences and so on.

    . . \begin{array}{cccccccccc}<br />
\text{Sequence} & \text{-}3 && \text{-}2 && 0 && 3 & \hdots \\<br />
\text{1st difference} && 1 && 2 && 3 \\<br />
\text{2nd difference} &&& 1 && 1 \end{array}


    We see that the second differences are constant.
    Hence, the generating function is of the second degree . . . a quadratic.

    The general quadratic function is: . f(n) \;=\;an^2 + bn + c


    Use the first three terms of the sequence and set up a system of equations:

    . . \begin{array}{cccccc}<br />
f(1) = \text{-}3: & a + b + c &=& \text{-}3 & [1] \\<br />
f(2) = \text{-}2: & 4a + 2b + c &=& \text{-}2 & [2] \\<br />
f(3) = 0: & 9a + 3b + c &=& 0 & [3] \end{array}

    . . \begin{array}{ccccc}<br />
\text{Subtract [2] - [1]:} & 3a + b &=& 1 & [4] \\<br />
\text{Subtract [3] - [2]:} & 5a + b &=& 2 & [5] \end{array}

    . . \begin{array}{cccccccc}<br />
\text{Subtract [5] - [4]:} & 2a \:=\: 1 & \Rightarrow & a \:=\:\frac{1}{2} \end{array}

    . . \begin{array}{ccccccc}\text{Substitute into [4]:} & 3(\frac{1}{2}) + b \:=\:1 & \Rightarrow & b \:=\:\text{-}\frac{1}{2} \end{array}

    . . \begin{array}{ccccccc}\text{Substitute into [1]:} & \frac{1}{2} - \frac{1}{2} + c \:=\:\text{-}3 & \Rightarrow & c \:=\:\text{-}3 \end{array}


    Therefore, the generating function is:

    . . . f(n) \;=\;\frac{1}{2}n^2 - \frac{1}{2}n - 3 \;=\;\dfrac{(n-3)(n+2)}{2}
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    Told you he would!!
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