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Math Help - Arithmetic series

  1. #1
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    Arithmetic series

    a simple one but I cant remember how to do this!!!


    Find sum of the series...

    <br />
\sum\limits_{r = 5}^{r = 90} {(3r - 2)} <br />
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  2. #2
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    Quote Originally Posted by dankelly07 View Post
    a simple one but I cant remember how to do this!!!


    Find sum of the series...

    <br />
\sum\limits_{r = 5}^{r = 90} {(3r - 2)} <br />
    \sum_{r = 5}^{90} {(3r - 2)}

    = \sum_{r = 5}^{90}3r + \sum_{r = 5}^{90}(-2)

    = 3 \sum_{r = 5}^{90}r - 2 \sum_{r = 5}^{90}1

    = 3 \left ( \sum_{r = 1}^{90}r - \sum_{r = 1}^4r \right ) - 2 \left ( \sum_{r = 1}^{90}1 - \sum_{r = 1}^41 \right )

    Can you take it from here?

    -Dan
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  3. #3
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    Hello,

    Quote Originally Posted by dankelly07 View Post
    a simple one but I cant remember how to do this!!!


    Find sum of the series...

    <br />
\sum\limits_{r = 5}^{r = 90} {(3r - 2)} <br />
    Let a_r=3r-2

    a_{r+1}=3(r+1)-2=(3r-2)+3=a_r+{\color{red}3}

    Therefore :

    \sum a_r is an arithmetic series of progression 3.

    Hmmm, can you continue ?


    --------------------

    Here is another method :

    \sum_{r=5}^{90} (3r-2)=3 \sum_{r=5}^{90} r-\sum_{r=5}^{90} 2

    You should know the general formula for \sum r.

    And for \sum 2, represent yourself adding 2, adding again, and again, and again..

    (let's see first if you can do the first one ^^)



    Edit : oh, well...topsquark spoiled it
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  4. #4
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    Hello, dankelly07!

    Another approach . . .


    Find sum of the series: . \sum\limits_{r = 5}^{90} (3r - 2)

    We have: . S \:=\:13 + 16 + 19 + 22 + \hdots + 268

    This is an arithmetic series.
    . . It has: .first term a = 13, common difference d = 3, and n = 86 terms.

    Its sum is: . S \;=\;\frac{n}{2}\bigg[2a + (n-1)d\bigg] \;=\;\frac{86}{2}\bigg[2(13) + 85(3)\bigg] \;=\;12,083

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  5. #5
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    thanks alot, all answers helped..
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