Question

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

Answer 1

@jim_thompson5910

Answer 2

attach the file by clicking "attach file" under the text box

Answer 3

thnx :)

Answer 4

np

Answer 5

i know the partial sum formula is used

Answer 6

so you'll use the formula

\[\Large \sum_{i=1}^{n} a*r^{n-1} = a*\frac{1-r^n}{1-r}\]

Answer 7

yes but i am finding difficult to write it down properly how ssould i obtain e

Answer 8

sorry I should have used i instead of n in the exponent

\[\Large \sum_{i=1}^{n} a*r^{\color{red}{\Huge i}-1} = a*\frac{1-r^n}{1-r}\]

Answer 9

let me think

Answer 10

so how would i get ratio , i think its e^(1/n)

Answer 11

sure :)

Answer 12

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?

Answer 13

the first term is e^1/n and the common ratio also e^1/n

Answer 14

`the first term is e^1/n`
correct

Answer 15

`common ratio also e^1/n`
incorrect

Answer 16

pick any term you want
divide it by the previous term

Answer 17

so for example

\[\Large r = \frac{e^{2/n}}{e^{1/n}} = ??\]

Answer 18

e^1/n

Answer 19

if u minus 2/n -1/n you get 1/n

Answer 20

oh sorry I was thinking of something else, yes you are correct

Answer 21

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

Answer 22

lucky me :)

Answer 23

now plug that into

\[\Large a*\frac{1-r^n}{1-r}\]

Answer 24

ok \[\frac{ e ^{1/n} (1-e) }{ (1-e ^{1/n})}\]

Answer 25

thats it

Answer 26

exactly
\[\Large a*\frac{1-r^n}{1-r}\]
\[\Large e^{1/n}*\frac{1-\left(e^{1/n}\right)^n}{1-e^{1/n}}\]
\[\Large e^{1/n}*\frac{1-e^{(1/n)*(n/1)}}{1-e^{1/n}}\]
\[\Large e^{1/n}*\frac{1-e^{n/n}}{1-e^{1/n}}\]
\[\Large e^{1/n}*\frac{1-e^{1}}{1-e^{1/n}}\]
\[\Large \frac{e^{1/n}(1-e)}{1-e^{1/n}}\]

Answer 27

thank you so much :)

Answer 28

no problem