anonymous
  • anonymous
hey everyone i have a question related to advance geometric series , mind giving it a try plz :)
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
@jim_thompson5910
jim_thompson5910
  • jim_thompson5910
attach the file by clicking "attach file" under the text box
anonymous
  • anonymous
thnx :)

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

jim_thompson5910
  • jim_thompson5910
np
anonymous
  • anonymous
i know the partial sum formula is used
jim_thompson5910
  • jim_thompson5910
so you'll use the formula \[\Large \sum_{i=1}^{n} a*r^{n-1} = a*\frac{1-r^n}{1-r}\]
anonymous
  • anonymous
yes but i am finding difficult to write it down properly how ssould i obtain e
jim_thompson5910
  • jim_thompson5910
sorry I should have used i instead of n in the exponent \[\Large \sum_{i=1}^{n} a*r^{\color{red}{\Huge i}-1} = a*\frac{1-r^n}{1-r}\]
jim_thompson5910
  • jim_thompson5910
let me think
anonymous
  • anonymous
so how would i get ratio , i think its e^(1/n)
anonymous
  • anonymous
sure :)
jim_thompson5910
  • jim_thompson5910
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?
anonymous
  • anonymous
the first term is e^1/n and the common ratio also e^1/n
jim_thompson5910
  • jim_thompson5910
`the first term is e^1/n` correct
jim_thompson5910
  • jim_thompson5910
`common ratio also e^1/n` incorrect
jim_thompson5910
  • jim_thompson5910
pick any term you want divide it by the previous term
jim_thompson5910
  • jim_thompson5910
so for example \[\Large r = \frac{e^{2/n}}{e^{1/n}} = ??\]
anonymous
  • anonymous
e^1/n
anonymous
  • anonymous
if u minus 2/n -1/n you get 1/n
jim_thompson5910
  • jim_thompson5910
oh sorry I was thinking of something else, yes you are correct
jim_thompson5910
  • jim_thompson5910
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
anonymous
  • anonymous
lucky me :)
jim_thompson5910
  • jim_thompson5910
now plug that into \[\Large a*\frac{1-r^n}{1-r}\]
anonymous
  • anonymous
ok \[\frac{ e ^{1/n} (1-e) }{ (1-e ^{1/n})}\]
anonymous
  • anonymous
thats it
jim_thompson5910
  • jim_thompson5910
exactly \[\Large a*\frac{1-r^n}{1-r}\] \[\Large e^{1/n}*\frac{1-\left(e^{1/n}\right)^n}{1-e^{1/n}}\] \[\Large e^{1/n}*\frac{1-e^{(1/n)*(n/1)}}{1-e^{1/n}}\] \[\Large e^{1/n}*\frac{1-e^{n/n}}{1-e^{1/n}}\] \[\Large e^{1/n}*\frac{1-e^{1}}{1-e^{1/n}}\] \[\Large \frac{e^{1/n}(1-e)}{1-e^{1/n}}\]
anonymous
  • anonymous
thank you so much :)
jim_thompson5910
  • jim_thompson5910
no problem

Looking for something else?

Not the answer you are looking for? Search for more explanations.