## anonymous one year ago use the formula to evaluate the series 3-9+27-81+...-a8

1. jim_thompson5910

2. triciaal

there are 2 types of series arithmetic which has a common difference from one term to the next and the geometric progression that has a common ratio from on term to the next. can you now identify what kind of series this is?

3. anonymous

i dont know how to do screen shot

4. anonymous

5. jim_thompson5910

thanks

6. anonymous

okay now i know

7. jim_thompson5910

each term is found by multiplying the previous term by -3

8. jim_thompson5910

3*(-3) = -9 -9*(-3) = 27 27*(-3) = -81 etc etc

9. anonymous

yeah

10. jim_thompson5910

generate the first 8 terms, then add them up

11. anonymous

okay hold on lets see

12. anonymous

-4920

13. jim_thompson5910

alternatively, you can use this formula $\Large S_n = a*\frac{1-r^n}{1-r}$

14. anonymous

yeah thats what its says to use

15. jim_thompson5910

yeah it's -4920

16. anonymous

okay lets see ill send you a screen shot if its wrong okay

17. anonymous

you are amazing

18. jim_thompson5910

ok I'm glad that one worked

19. anonymous

okay i have another

20. anonymous

i have a question are you a teacher for algebra

21. anonymous

because you seem to know what your doing

22. jim_thompson5910

yes I've had lots and lots of practice

23. anonymous

sweet heres the other question

24. anonymous

25. jim_thompson5910

this is a geometric series what's the first term? common ratio?

26. anonymous

1/3

27. jim_thompson5910

first term = 1/3, yes

28. anonymous

its going up in numerator x2 denomanator x3

29. jim_thompson5910

common ratio = -2/3

30. anonymous

okay

31. anonymous

so its diverge

32. anonymous

right

33. jim_thompson5910

r = -2/3 since |r| < 1 is true, this means the infinite series does converge and there is a fixed sum it reaches that sum would be S = a/(1-r) S = (1/3)/(1-(-2/3)) S = (1/3)/(5/3) S = (1/3)*(3/5) S = 1/5

34. jim_thompson5910

Rule: if |r| < 1, then the series converges otherwise, the series diverges

35. anonymous

okay so its fixed sum is 1/5

36. jim_thompson5910

if you generated all of the terms and added them up, they would add up to 1/5

37. anonymous

okay

38. anonymous

so converges are out right

39. jim_thompson5910

what do you mean? I wrote above that the series converges

40. jim_thompson5910

since |r|<1