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Math Help - Arithmetic Series

  1. #1
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    Arithmetic Series

    So there is wall...the top has 1 brick, then the next row has 3 bricks, then 5 bricks etc. there are 2500 bricks in the wall then how many rows are there

    im just really confused with arithmetic series right now and need general help with them too....

    pref. without using some formula but a more explanative version

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  2. #2
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    Quote Originally Posted by stones44 View Post
    So there is wall...the top has 1 brick, then the next row has 3 bricks, then 5 bricks etc. there are 2500 bricks in the wall then how many rows are there

    im just really confused with arithmetic series right now and need general help with them too....

    pref. without using some formula but a more explanative version

    This would be a series of the odd numbers so it would be \sum_{i=0}^{n}2i+1

    can you find n from that?
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  3. #3
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    no thats what im saying..im very confused
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Mathstud29 View Post
    This would be a series of the odd numbers so it would be \sum_{i=0}^{n}2i+1

    can you find n from that?
    \sum_{n=0}^{x}2n+1=\sum_{n=0}^{x}2n+\sum_{n=0}^{x}  1=\frac{2x(x+1)}{2}+x

    Now solve \frac{2x(x+1)}{2}+x=2500
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  5. #5
    Super Member flyingsquirrel's Avatar
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    Hi
    Quote Originally Posted by Mathstud28 View Post
    \sum_{n=0}^{x}2n+1=\sum_{n=0}^{x}2n+\sum_{n=0}^{x}  1=\frac{2x(x+1)}{2}+x{\color{red}+1}
    There are n-p+1 elements in the set of positive integers \{p,\,p+1,\,p+2,\,\ldots,\,n-1,\,n\}, assuming n\geq p.
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  6. #6
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    Hello, stones44!

    So there is wall.
    The top has 1 brick, then the next row has 3 bricks, then 5 bricks, etc.
    There are 2500 bricks in the wall.
    How many rows are there?

    Without a formula? . . . Then a child can do it!

    Just start adding: . 1 + 3 + 5 + 7 + 9 + \hdots . until you reach 2500.

    Then count up the number of numbers you added.

    . . . I'll wait in the car . . .



    Formula for the sum of the first n terms of an arithmetic series:
    . . S_n \;=\;\frac{n }{2}[2a + (n-1)d]

    We have: .first term, a = 1 . . . common difference, d = 2

    We know: . S_n = 2500 . . . and we want n.


    So we have: . 2500 \:=\:\frac{n}{2}[2(1) + (n-1)2] \quad\Rightarrow\quad 2500 \:=\:\frac{n}{2}(2n)\quad\Rightarrow\quad n^2 \:=\:2500

    Therefore: . n \:=\:50 . . . There are 50 rows.

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