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Math Help - geometric progression

  1. #1
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    geometric progression

    Hello
    Can someone explain why multiplying the second formula by 'r' in the derivation of this formula doesn't change things in any way before subtracting?
    ...or just some background on the formula
    Thanks
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  2. #2
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    Quote Originally Posted by 200001 View Post
    Hello
    Can someone explain why multiplying the second formula by 'r' in the derivation of this formula doesn't change things in any way before subtracting?
    ...or just some background on the formula
    Thanks
    It does change things slightly. You had something like 1+r+r^2+r^3+r^4+....+r^n

    when you multiply by r you get r+r^2+r^3+r^4+...+r^n+r^{n+1}

    see how the we no longer have the 1? but we picked up that last term, then you subtract
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  3. #3
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    proof of sum of geometric series

    I guess I should make this formal to explain it better

    If we are dealing with finite series we have this

    \displaystyle\sum_{k=0}^n ar^k=s=a+ar+ar^2+...+ar^n

    So that means sr=ar+ar^2+...+ar^{n+1}

    So now lets subtract the two

    s-sr=(a+ar+ar^2+...+ar^n)-(ar+ar^2+...+ar^{n+1})

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

    So s-sr=s(1-r)=a(1-r^{n+1})

    And so s=\frac{a(1-r^{n+1})}{1-r}

    So multiplying by r did change something, we didnt have the 1 and we picked up a r^{n+1}

    Now what if the series is infinite? Just take the limit as n goes to infinity. If |r|<1 then the r^{n+1} tends to zero and so the formula becomes \frac{a}{1-r}
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  4. #4
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    Thanks very much for taking the time to write that. It have been a great help

    Now, if your series have started at 1 instead of 0 would that then leave use with R^n in the final summation in the numerator instead of the R^n+1 ?

    Again, thanks for this
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  5. #5
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    Quote Originally Posted by 200001 View Post
    Thanks very much for taking the time to write that. It have been a great help

    Now, if your series have started at 1 instead of 0 would that then leave use with R^n in the final summation in the numerator instead of the R^n+1 ?

    Again, thanks for this
     <br />
\displaystyle\sum_{k=1}^n ar^k=s=ar+ar^2+...+ar^n<br />

    sr=ar^2+ar^3+...+ar^{n+1}

    s-sr=s(1-r)=ar-ar^{n+1}

    s=\frac{a(r-r^{n+1})}{1-r}

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