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Math Help - Formula

  1. #1
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    Formula

    Can someone tell me the formula for finding the sum of a finite qeometric series
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  2. #2
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    let S=\sum\limits_{i=1}^{n}ar^{i-1}

    then
    S=a+ar+ar^{2}+\cdots +ar^{n-1}

    and

    rS=ar+ar^{2}+ar^{3}+\cdots +ar^{n}

    substract this two equations yield

    S(1-r)=a-ar^{n}

    S=\frac{a-ar^{n}}{(1-r)}
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  3. #3
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    Quote Originally Posted by dedust View Post
    let S=\sum\limits_{i=1}^{n}ar^{i-1}

    then
    S=a+ar+ar^{2}+\cdots +ar^{n-1}

    and

    rS=ar+ar^{2}+ar^{3}+\cdots +ar^{n}

    substract this two equations yield

    S(1-r)=a-ar^{n}

    S=\frac{a-ar^{n}}{(1-r)}


    How do u find n
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  4. #4
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    Quote Originally Posted by Tinhorn View Post
    How do u find n
    HI

    Consider this series : 1+2+3+4

    the n here would be 4

    is this what you want toknow ?
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  5. #5
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    thank you
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  6. #6
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    Quote Originally Posted by mathaddict View Post
    HI

    Consider this series : 1+2+3+4

    the n here would be 4

    is this what you want toknow ?

    would that logic also worked for 1 + 2/3 + 4/9 +........ + 64/729
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  7. #7
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    Quote Originally Posted by Tinhorn View Post
    would that logic also worked for 1 + 2/3 + 4/9 +........ + 64/729
    n is the number of terms in the series.

    what is n for this series?
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  8. #8
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    Quote Originally Posted by dedust View Post
    n is the number of terms in the series.

    what is n for this series?
    That's what I'm trying to figure out
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  9. #9
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    look at the formula,.
    the first term a=1, the ratio r=2/3, and
    the last term is 64/729
    so

    ar^{n-1}=64/729
    \left(\frac{2}{3}\right)^{(n-1)}=\frac{64}{729}=\left(\frac{2}{3}\right)^6
    n-1=6
    n=7
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