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Math Help - Area of figure (increasing)

  1. #1
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    Area of figure (increasing)

    How do you do part (c)? Is there a shorter way than calculating everything upto 100?
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  2. #2
    MHF Contributor Also sprach Zarathustra's Avatar
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    Re: Area of figure (increasing)

    Quote Originally Posted by fleurdel View Post
    How do you do part (c)? Is there a shorter way than calculating everything upto 100?

    100*100+2+101+1
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  3. #3
    Senior Member BAdhi's Avatar
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    Re: Area of figure (increasing)

    A_2=A_1+2(1)
    A_3=A_2+2(2)
    A_4=A_3+2(3)
    \vdots
    A_{n-1}=A_{n-2}+2(n-2)
    A_{n}=A_{n-1}+2(n-1)

    add downwards, which eventually gives

    A_n=A_1+2(1+2+3+...+n-1)

    take A_1=4

    A_n=4+2\sum_{r=1}^{n-1} r
    Last edited by BAdhi; July 3rd 2011 at 02:15 AM. Reason: made more clear
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  4. #4
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    Re: Area of figure (increasing)

    Hello, fleurdel!

    How do you do part (c)?
    Is there a shorter way than calculating everything up to 100?

    (c) Find the area of Figure 100.

    . . \begin{array}{cc} \text{Figure} & \text{Area} \\ \hline 1 & 6 \\ 2 & 10 \\ 3 & 16 \\ 4 & 24 \\ \vdots & \vdots \end{array}

    We have this sequence: . 6,\,10,\,16,\,24\, \hdots

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

    . . \begin{array}{cccccccc}\text{Sequence} & 6 && 10 && 16 && 24 \\ \text{1st diff.} && 4 && 6 && 8 \\ \text{2nd diff.} &&& 2 && 2 \end{array}

    We see that the second differences are constant.
    This mean that 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 to construct a system of equations:

    . . \begin{array}{cccccc}f(1) = 6 & a + b + c &=& 6 & [1] \\ f(2) = 6 & 4a + 2b + c &=& 10 & [2] \\ f(3) = 16 & 9a + 3b + c &=& 16 & [3] \end{array}

    . . \begin{array}{cccccc}\text{Subtract [2] - [1]:} & 3a + b &=& 4 & [4] \\ \text{Subtract [3] - [2]:} & 5a + b &=& 6 & [5] \end{array}

    . . \text{Subtract [5] - [4]: }\;2a \:=\:2 \quad\Rightarrow\quad a \:=\:1

    \text{Substitute into [4]: }\:3 + b \:=\:4 \quad\Rightarrow\quad b \:=\:1

    \text{Substitute into [1]: }\; 1 + 1 + c \:=\:6 \quad\Rightarrow\quad c \:=\:4

    . . \text{Hence: }\:f(n) \:=\:n^2 + n + 4


    \text{Therefore: }\:f(100) \:=\:100^2 + 100 + 4 \:=\:\boxed{10,\!104}

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