## anonymous one year ago hey everyone i have a question related to advance geometric series , mind giving it a try plz :)

1. anonymous

@jim_thompson5910

2. jim_thompson5910

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

3. anonymous

thnx :)

4. jim_thompson5910

np

5. anonymous

i know the partial sum formula is used

6. jim_thompson5910

so you'll use the formula $\Large \sum_{i=1}^{n} a*r^{n-1} = a*\frac{1-r^n}{1-r}$

7. anonymous

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

8. jim_thompson5910

sorry I should have used i instead of n in the exponent $\Large \sum_{i=1}^{n} a*r^{\color{red}{\Huge i}-1} = a*\frac{1-r^n}{1-r}$

9. jim_thompson5910

let me think

10. anonymous

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

11. anonymous

sure :)

12. 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?

13. anonymous

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

14. jim_thompson5910

the first term is e^1/n correct

15. jim_thompson5910

common ratio also e^1/n incorrect

16. jim_thompson5910

pick any term you want divide it by the previous term

17. jim_thompson5910

so for example $\Large r = \frac{e^{2/n}}{e^{1/n}} = ??$

18. anonymous

e^1/n

19. anonymous

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

20. jim_thompson5910

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

21. 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

22. anonymous

lucky me :)

23. jim_thompson5910

now plug that into $\Large a*\frac{1-r^n}{1-r}$

24. anonymous

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

25. anonymous

thats it

26. jim_thompson5910

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

27. anonymous

thank you so much :)

28. jim_thompson5910

no problem