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

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

Mathematics
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thnx :)

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Other answers:

np
i know the partial sum formula is used
so you'll use the formula \[\Large \sum_{i=1}^{n} a*r^{n-1} = a*\frac{1-r^n}{1-r}\]
yes but i am finding difficult to write it down properly how ssould i obtain e
sorry I should have used i instead of n in the exponent \[\Large \sum_{i=1}^{n} a*r^{\color{red}{\Huge i}-1} = a*\frac{1-r^n}{1-r}\]
let me think
so how would i get ratio , i think its e^(1/n)
sure :)
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?
the first term is e^1/n and the common ratio also e^1/n
`the first term is e^1/n` correct
`common ratio also e^1/n` incorrect
pick any term you want divide it by the previous term
so for example \[\Large r = \frac{e^{2/n}}{e^{1/n}} = ??\]
e^1/n
if u minus 2/n -1/n you get 1/n
oh sorry I was thinking of something else, yes you are correct
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
lucky me :)
now plug that into \[\Large a*\frac{1-r^n}{1-r}\]
ok \[\frac{ e ^{1/n} (1-e) }{ (1-e ^{1/n})}\]
thats it
exactly \[\Large a*\frac{1-r^n}{1-r}\] \[\Large e^{1/n}*\frac{1-\left(e^{1/n}\right)^n}{1-e^{1/n}}\] \[\Large e^{1/n}*\frac{1-e^{(1/n)*(n/1)}}{1-e^{1/n}}\] \[\Large e^{1/n}*\frac{1-e^{n/n}}{1-e^{1/n}}\] \[\Large e^{1/n}*\frac{1-e^{1}}{1-e^{1/n}}\] \[\Large \frac{e^{1/n}(1-e)}{1-e^{1/n}}\]
thank you so much :)
no problem

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