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Mathematics 15 Online
OpenStudy (anonymous):

use the formula to evaluate the series 3-9+27-81+...-a8

jimthompson5910 (jim_thompson5910):

post the screenshot please

OpenStudy (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?

OpenStudy (anonymous):

i dont know how to do screen shot

OpenStudy (anonymous):

jimthompson5910 (jim_thompson5910):

thanks

OpenStudy (anonymous):

okay now i know

jimthompson5910 (jim_thompson5910):

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

jimthompson5910 (jim_thompson5910):

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

OpenStudy (anonymous):

yeah

jimthompson5910 (jim_thompson5910):

generate the first 8 terms, then add them up

OpenStudy (anonymous):

okay hold on lets see

OpenStudy (anonymous):

-4920

jimthompson5910 (jim_thompson5910):

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

OpenStudy (anonymous):

yeah thats what its says to use

jimthompson5910 (jim_thompson5910):

yeah it's -4920

OpenStudy (anonymous):

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

OpenStudy (anonymous):

you are amazing

jimthompson5910 (jim_thompson5910):

ok I'm glad that one worked

OpenStudy (anonymous):

okay i have another

OpenStudy (anonymous):

i have a question are you a teacher for algebra

OpenStudy (anonymous):

because you seem to know what your doing

jimthompson5910 (jim_thompson5910):

yes I've had lots and lots of practice

OpenStudy (anonymous):

sweet heres the other question

OpenStudy (anonymous):

jimthompson5910 (jim_thompson5910):

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

OpenStudy (anonymous):

1/3

jimthompson5910 (jim_thompson5910):

first term = 1/3, yes

OpenStudy (anonymous):

its going up in numerator x2 denomanator x3

jimthompson5910 (jim_thompson5910):

common ratio = -2/3

OpenStudy (anonymous):

okay

OpenStudy (anonymous):

so its diverge

OpenStudy (anonymous):

right

jimthompson5910 (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

jimthompson5910 (jim_thompson5910):

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

OpenStudy (anonymous):

okay so its fixed sum is 1/5

jimthompson5910 (jim_thompson5910):

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

OpenStudy (anonymous):

okay

OpenStudy (anonymous):

so converges are out right

jimthompson5910 (jim_thompson5910):

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

jimthompson5910 (jim_thompson5910):

since |r|<1

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