Jadzia
  • Jadzia
I need help. Question below. The answer is e-1, and I have no idea how they got that.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Jadzia
  • Jadzia
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Jadzia
  • Jadzia
@dan815 @tHe_FiZiCx99 @mathstudent55
Jadzia
  • Jadzia
@Luigi0210

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anonymous
  • anonymous
Try over here: https://mathway.com/
anonymous
  • anonymous
Otherwise, I don't know how to do it.
anonymous
  • anonymous
This is a geometric series. It may be easier to write it as follows: \[ \frac{1}{n} \sum_{k=1}^n \left(e^{\frac{1}{n}}\right)^k\]
Jadzia
  • Jadzia
i got an idea.. probably f(x)=e^x and then delta x is 1/n |dw:1450074305872:dw|
Jadzia
  • Jadzia
oh wait nvm ... hmm let me think xD
bubblegum.
  • bubblegum.
when you write the whole thing it will be like this-> \[\Large\frac{ 1 }{ n} \color{red}{(e^\frac{ 1 }{ n }+e^\frac{ 2 }{ n }+e^\frac{ 3 }{ n }+....+e^\frac{ n }{ n })}\] now the whole thing inside the bracket is a Geometric sequence :D with the common ratio of \(\Large\color{blue}{e^\frac{1}{n}}\) and the summation till \(n^{th}\) term of a Geometric sequence is given by->\(\Large\color{gold}{\frac{ t_1(1-r^n) }{ 1-r }}\) here \(t_1\) is the 1st term of the GP(Geometric sequence) \(r\) is the common ratio \(n\) is the number of terms in the summation so we have this GP whos sum we have to find-> \(\Large\color{red}{(e^\frac{ 1 }{ n }+e^\frac{ 2 }{ n }+e^\frac{ 3 }{ n }+....+e^\frac{ n }{ n })}\) \(t_1\)=\(e^{\frac{1}{n}}\) \(r\)=\(e^{\frac{1}{n}}\) now just find this summation and put the value in our equation \[\Large\frac{ 1 }{ n} \color{red}{(e^\frac{ 1 }{ n }+e^\frac{ 2 }{ n }+e^\frac{ 3 }{ n }+....+e^\frac{ n }{ n })}\] :)
Jadzia
  • Jadzia
I think i got it, Reimann sum :) thanks!
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Jadzia
  • Jadzia
**ignore the last part, i just tested something out**
bubblegum.
  • bubblegum.
hmm according to what i did i got this-> \[\frac{ 1 }{ n }\left( \frac{ e^\frac{ 1 }{ n } (1-e)}{ 1-e^\frac{ 1 }{ n } } \right)\] and then i took n=1 to check if it comes equal to e-1 but its not when we take n=1 then frm this formula we get the answer as ->e and also if you put n=1 in our original equation you get answer as e but ur book says it is e-1 :/
bubblegum.
  • bubblegum.
http://www.wolframalpha.com/input/?i=%5Cfrac%7B1%7D%7Bn%7D+%5Csum_%7Bk%3D1%7D%5E%7Bn%7De%5E%7Bk%2Fn%7D
Jadzia
  • Jadzia
hmm i think you should apply the limit there..|dw:1450075474500:dw||dw:1450075597451:dw||dw:1450075754882:dw|
bubblegum.
  • bubblegum.
oh so n must be considered as \(\infty\) ??
Jadzia
  • Jadzia
n represents the number of terms and we are looking for the infinite value for n, not the 1st term. (not sure about the proper wording, but hope you get what i mean :P )
bubblegum.
  • bubblegum.
oh :O okay :)
Jadzia
  • Jadzia
:) g'night

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