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Thread: 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 $\displaystyle S=\sum\limits_{i=1}^{n}ar^{i-1}$

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

    and

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

    substract this two equations yield

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

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

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

    and

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

    substract this two equations yield

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

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


    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
    $\displaystyle n$ 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
    $\displaystyle n$ 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 $\displaystyle a=1$, the ratio $\displaystyle r=2/3$, and
    the last term is $\displaystyle 64/729$
    so

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