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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 :)
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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jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.4attach the file by clicking "attach file" under the text box

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i know the partial sum formula is used

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.4so you'll use the formula \[\Large \sum_{i=1}^{n} a*r^{n1} = a*\frac{1r^n}{1r}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes but i am finding difficult to write it down properly how ssould i obtain e

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.4sorry I should have used i instead of n in the exponent \[\Large \sum_{i=1}^{n} a*r^{\color{red}{\Huge i}1} = a*\frac{1r^n}{1r}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so how would i get ratio , i think its e^(1/n)

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.4let'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?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the first term is e^1/n and the common ratio also e^1/n

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.4`the first term is e^1/n` correct

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.4`common ratio also e^1/n` incorrect

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.4pick any term you want divide it by the previous term

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.4so for example \[\Large r = \frac{e^{2/n}}{e^{1/n}} = ??\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0if u minus 2/n 1/n you get 1/n

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.4oh sorry I was thinking of something else, yes you are correct

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.4so 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

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.4now plug that into \[\Large a*\frac{1r^n}{1r}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok \[\frac{ e ^{1/n} (1e) }{ (1e ^{1/n})}\]

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.4exactly \[\Large a*\frac{1r^n}{1r}\] \[\Large e^{1/n}*\frac{1\left(e^{1/n}\right)^n}{1e^{1/n}}\] \[\Large e^{1/n}*\frac{1e^{(1/n)*(n/1)}}{1e^{1/n}}\] \[\Large e^{1/n}*\frac{1e^{n/n}}{1e^{1/n}}\] \[\Large e^{1/n}*\frac{1e^{1}}{1e^{1/n}}\] \[\Large \frac{e^{1/n}(1e)}{1e^{1/n}}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0thank you so much :)
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