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anonymous

  • one year ago

hey everyone i have a question related to advance geometric series , mind giving it a try plz :)

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  1. anonymous
    • one year ago
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    @jim_thompson5910

  2. jim_thompson5910
    • one year ago
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    attach the file by clicking "attach file" under the text box

  3. anonymous
    • one year ago
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    thnx :)

  4. jim_thompson5910
    • one year ago
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    np

  5. anonymous
    • one year ago
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    i know the partial sum formula is used

  6. jim_thompson5910
    • one year ago
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    so you'll use the formula \[\Large \sum_{i=1}^{n} a*r^{n-1} = a*\frac{1-r^n}{1-r}\]

  7. anonymous
    • one year ago
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    yes but i am finding difficult to write it down properly how ssould i obtain e

  8. jim_thompson5910
    • one year ago
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    sorry I should have used i instead of n in the exponent \[\Large \sum_{i=1}^{n} a*r^{\color{red}{\Huge i}-1} = a*\frac{1-r^n}{1-r}\]

  9. jim_thompson5910
    • one year ago
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    let me think

  10. anonymous
    • one year ago
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    so how would i get ratio , i think its e^(1/n)

  11. anonymous
    • one year ago
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    sure :)

  12. jim_thompson5910
    • one year ago
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    let's write out the few terms of this summation \[\LARGE \sum_{i=1}^{n} e^{i/n} = e^{1/n}+e^{2/n}+e^{3/n}+\ldots e^{n/n}\] what's the first term? what's the common ratio?

  13. anonymous
    • one year ago
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    the first term is e^1/n and the common ratio also e^1/n

  14. jim_thompson5910
    • one year ago
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    `the first term is e^1/n` correct

  15. jim_thompson5910
    • one year ago
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    `common ratio also e^1/n` incorrect

  16. jim_thompson5910
    • one year ago
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    pick any term you want divide it by the previous term

  17. jim_thompson5910
    • one year ago
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    so for example \[\Large r = \frac{e^{2/n}}{e^{1/n}} = ??\]

  18. anonymous
    • one year ago
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    e^1/n

  19. anonymous
    • one year ago
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    if u minus 2/n -1/n you get 1/n

  20. jim_thompson5910
    • one year ago
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    oh sorry I was thinking of something else, yes you are correct

  21. jim_thompson5910
    • one year ago
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    so a = e^(1/n) is the first term r = e^(1/n) is the common ratio they coincidentally are the same expression, but this isn't always the case

  22. anonymous
    • one year ago
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    lucky me :)

  23. jim_thompson5910
    • one year ago
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    now plug that into \[\Large a*\frac{1-r^n}{1-r}\]

  24. anonymous
    • one year ago
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    ok \[\frac{ e ^{1/n} (1-e) }{ (1-e ^{1/n})}\]

  25. anonymous
    • one year ago
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    thats it

  26. jim_thompson5910
    • one year ago
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    exactly \[\Large a*\frac{1-r^n}{1-r}\] \[\Large e^{1/n}*\frac{1-\left(e^{1/n}\right)^n}{1-e^{1/n}}\] \[\Large e^{1/n}*\frac{1-e^{(1/n)*(n/1)}}{1-e^{1/n}}\] \[\Large e^{1/n}*\frac{1-e^{n/n}}{1-e^{1/n}}\] \[\Large e^{1/n}*\frac{1-e^{1}}{1-e^{1/n}}\] \[\Large \frac{e^{1/n}(1-e)}{1-e^{1/n}}\]

  27. anonymous
    • one year ago
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    thank you so much :)

  28. jim_thompson5910
    • one year ago
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    no problem

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