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Math Help - HELP on grade 11 math problem on sequences and series

  1. #1
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    HELP on grade 11 math problem on sequences and series

    i am totally confused about this practice problem. what steps are required i feel it's so overwhelming where do i start.

    At a grocery store oranges are piled to form a pyramid shape. the person who stacks the fruit has found that 18 by 26 oranges will fit in the bottom row. each successive row can can fit one less orange in both the length and width. if this pattern continues, how many oranges will there be in the pyramid? (it is important you know how many rows the pyramid will have.)
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  2. #2
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    You're being asked to work out

    18 \times 26 + 17 \times 25 + 16 \times 24 + \dots.

    How many terms do you think you're going to have?
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  3. #3
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    i still don't understand.. -_- what would be the formula??
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  4. #4
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    This would be the formula, I guess:
    \displaystyle\sum\limits_{n=1}^{18} n(n+8)

    =\displaystyle\sum\limits_{n=1}^{18} (n^2 + 8n)

    =\displaystyle\sum\limits_{n=1}^{18} n^2 + \displaystyle\sum\limits_{n=1}^{18} 8n

    =\displaystyle\sum\limits_{n=1}^{18} n^2 + 8\displaystyle\sum\limits_{n=1}^{18} n

    Note that \displaystyle\sum\limits_{n=1}^{k} n^2 = \frac{k(k + 1)(2k + 1)}{6} and \displaystyle\sum\limits_{n=1}^{k} n = \frac{k^2 + k}{2}:

    =\displaystyle\sum\limits_{n=1}^{18} n^2 + 8\displaystyle\sum\limits_{n=1}^{18} n

    =\frac{18(18 + 1)(2 \cdot 18 + 1)}{6} + 8 \left( \frac{18^2 + 18}{2}\right)

    =3477
    Last edited by eumyang; July 26th 2010 at 07:52 PM.
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  5. #5
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    -_- i don't understand this..i never learned this in grade 11 i think we have to use one of the arithmetic/geometric sequences and series
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  6. #6
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    Well, why don't you try it with a simpler example; say 4 by 6 in bottom row.
    The next row will be 3 by 5, then 2 by 4, finally 1 by 3 at top : still with me?

    So from top down, you have 3, 8, 15, 24 ; a total of 50, right?
    Notice that these go up by 5, then 7, then 9.
    Can you handle that with what you've learned?
    If so, apply same strategy to your 18 by 26 case.
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  7. #7
    Grand Panjandrum
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    Quote Originally Posted by leafs50 View Post
    i still don't understand.. -_- what would be the formula??
    You are not looking for a formula here but for a method, look at the pattern in #2 by Prove_It.

    What he has is the sum of the number of fruits in each layer, just keep going until a layer has no fruits, then do the sum.

    CB
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  8. #8
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    Hello, leafs50!

    Your usual sequence/series formulas won't help you with this one.
    We can simply crank out the answer . . . or derive a formula.
    Neither method is pleasant.



    At a grocery store, oranges are piled to form a pyramid shape.
    The produce manager has found that 18 by 26 oranges will fit in the bottom level.
    Each successive level can can fit one less orange in both the length and width.

    If this pattern continues, how many oranges will there be in the pyramid?

    The no-brainer method:

    . . \begin{array}{c|c|c} \text{Level} & \text {Dimensions} & \text{Oranges} \\ \hline 1 & 18\times 26 & 468 \\ 2 & 17 \times 25 & 425 \\ 3 & 16 \times 24 & 384 \\ 4 & 15 \times 23 & 345 \\ \vdots & \vdots & \vdots \\ 16 & 3 \times 11 & 33 \\ 17 & 2 \times 10 & 20 \\ 18 & 1 \times 9 & 9 \\ \hline & \text{Total:} & 3477\end{array}


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    We can derive the formula, but it will take a while . . .


    Reading from the top of the pyramid, we have these facts:

    . . \begin{array}{c|c||c}<br />
\text{Level} & \text{Oranges} & \text{Total} \\ \hline 1 & 9 & 9 \\ 2 & 20 & 29 \\ 3 & 33 & 62 \\ 4 & 48 & 110 \\ 5 & 65 & 175 \\ 6 & 84 & 259 \\ \vdots & \vdots & \vdots\end{array}

    The Totals form this sequence: . 9,29,62,110,175,259, \hdots


    Take the difference of consecutive terms,
    . . then the differences of the differences, etc.


    \begin{array}{c|ccccccccccccc}<br />
\text{Sequence} &  9 && 29 && 62 && 110 && 175 && 259 \\ \hline<br />
\text{1st diff.} && 20 && 33 && 48 && 65 && 84 \\<br />
\text{2nd diff.} &&& 13 && 15 && 17 && 19 \\<br />
\text{3rd diff.} &&&& 2 && 2 && 2\end{array}


    We see that the third differences are constant.
    . . Hence, the generating function is of the third degree.

    The general cubic function is: . f(n) \:=\:an^3 + bn^2 + cn + d
    . . and we must determine a,b,c,d.


    We know the first four terms of the sequence:

    . . \begin{array}{cccccccccc}<br />
f(1) = 9: & a + b + c + d &=& 9 & [1] \\<br />
f(2) = 29: & 8a + 4b + 2c + d &=& 29 & [2] \\<br />
f(3) = 62: & 27a + 9b + 3c + d &=& 62 & [3] \\<br />
f(4) = 110: & 64a + 16b + 4c + d &=& 110 & [4] \end{array}


    \begin{array}{cccccc}<br />
\text{Subtract [2]-[1]:} & 7a + 3b + c &=& 20 & [5] \\<br />
\text{Subtract [3]-[2]:} & 19a + 5b + c &=& 33 & [6] \\<br />
\text{Subtract [4]-[3]:} & 37a + 7b + c &=& 48 & [7] \end{array}


    \begin{array}{cccccc}<br />
\text{Subtract [6]-[5]:} & 12a + 2b &=& 13 & [8] \\<br />
\text{Subtract [7]-[6]:} & 18a + 2b &=& 15 & [9] \end{array}


    \begin{array}{ccccccc}<br />
\text{Subtract [9]-[8]:} & 6a \;=\l2 & \Rightarrow & \boxed{a \:=\:\tfrac{1}{3}} \end{array}

    Substitute into [8]: . 12\left(\frac{1}{3}\right) + 2b \:=\:13 \quad\Rightarrow\quad \boxed{b \:=\:\tfrac{9}{2}}

    Substitute into [5]: . 7\left(\frac{1}{3}\right) + 3\left(\frac{9}{2}\right) + c \:=\:20 \quad\Rightarrow\quad \boxed{c \:=\:\tfrac{25}{6}}

    Substitute into [1]: . \frac{1}{3} + \frac{9}{2} + \frac{25}{6} + d \:=\:9 \quad\Rightarrow\quad \boxed{d \:=\:0}



    Hence, the generating function is: . f(n) \;=\;\frac{1}{3}n^3 + \frac{9}{2}n^2 + \frac{25}{6}n <br />


    Therefore: . f(n) \;=\;\frac{n}{6}(2n^2 + 27n + 25) \;=\;\dfrac{n(n+1)(2n+25)}{6}

    . . and: . f(18) \;=\;\dfrac{18(19)(61)}{6} \;=\;3477

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