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Math Help - Series

  1. #1
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    Series

    How would I write this series with the  \Sigma sign and all:

    25 (1 + 2.04 + 3.04 + 4.08 + 5.08 + 6.12 + 7.12 + 8.16 ...)

    OR

    25 + 51 + 76 + 102 + 127 ....
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  2. #2
    Senior Member TheAbstractionist's Avatar
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    \sum_{n\,=\,0}^k25\left(n+0.04\left\lfloor\frac n2\right\rfloor\right)

    =\quad\sum_{n\,=\,0}^k\left(25n+\left\lfloor\frac n2\right\rfloor\right)
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  3. #3
    Super Member Deadstar's Avatar
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    Right I see the abstractionist is also looking at this thread so by the time i post this he''ll have a better answer down probably...

    Do it as two sums...

    \bigg{(} \bigg{(} 25 \sum_{n=0}^\infty 2n \bigg{)} + n \bigg{)} + \bigg{(} \bigg{(} 25 \sum_{n=0}^\infty (2n+1) \bigg{)} + n \bigg{)}
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  4. #4
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    Hello, metlx!

    Write in sigma-notation:

    (a)\;\;S \;=\;25\,\bigg(1 + 2.04 + 3.04 + 4.08 + 5.08 + 6.12 + 7.12 + 8.16 + \hdots\bigg)
    This is quicky tricky . . .


    \text{We have: }\;S \;=\;\;25\bigg(1 + \underbrace{2.04 + 3.04} + \underbrace{4.08 + 5.08} + \underbrace{6.12 + 7.12} +  \underbrace{8.16 + 9.16} + \hdots\bigg)

    . . . . . . S \;=\;25\bigg(1 + 5.08 + 9.16 + 13.24 + 17.32 + \hdots\bigg)

    . . . . . . S \;=\;25\bigg([1 + 0(0.8)] + [5 + 1(0.08)] + [9 + 2(0.8)] + [13 + 3(0.8)] + \hdots\bigg)

    . . . . . . S \;=\;25\sum_{k=0}^n\bigg((4k+1) + k(0.8)\bigg)  \;=\;25\sum_{k=0}^n\bigg(4.8k + 1\bigg)




    (b)\;\;S \;=\;25 + 51 + 76 + 102 + 127 + \hdots
    This too is tricky. .The pattern is: add 26, add 25, add 26, add 25, . . .


    \text{We have: }\;S \;=\;\underbrace{25 + 51} + \underbrace{76 + 102} + \underbrace{127 + 153} + \underbrace{178 + 204} + \hdots

    . . . . . . S \;=\;76 + 178 + 280 + 382 + \hdots . which has a common difference of 102


    Therefore: . S \;=\;\sum_{k=0}^n\bigg(76 + 102k\bigg)

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