anonymous
  • anonymous
use the formula to evaluate the series 3-9+27-81+...-a8
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!
jim_thompson5910
  • jim_thompson5910
post the screenshot please
triciaal
  • 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?
anonymous
  • anonymous
i dont know how to do screen shot

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anonymous
  • anonymous
1 Attachment
jim_thompson5910
  • jim_thompson5910
thanks
anonymous
  • anonymous
okay now i know
jim_thompson5910
  • jim_thompson5910
each term is found by multiplying the previous term by -3
jim_thompson5910
  • jim_thompson5910
3*(-3) = -9 -9*(-3) = 27 27*(-3) = -81 etc etc
anonymous
  • anonymous
yeah
jim_thompson5910
  • jim_thompson5910
generate the first 8 terms, then add them up
anonymous
  • anonymous
okay hold on lets see
anonymous
  • anonymous
-4920
jim_thompson5910
  • jim_thompson5910
alternatively, you can use this formula \[\Large S_n = a*\frac{1-r^n}{1-r}\]
anonymous
  • anonymous
yeah thats what its says to use
jim_thompson5910
  • jim_thompson5910
yeah it's -4920
anonymous
  • anonymous
okay lets see ill send you a screen shot if its wrong okay
anonymous
  • anonymous
you are amazing
jim_thompson5910
  • jim_thompson5910
ok I'm glad that one worked
anonymous
  • anonymous
okay i have another
anonymous
  • anonymous
i have a question are you a teacher for algebra
anonymous
  • anonymous
because you seem to know what your doing
jim_thompson5910
  • jim_thompson5910
yes I've had lots and lots of practice
anonymous
  • anonymous
sweet heres the other question
anonymous
  • anonymous
1 Attachment
jim_thompson5910
  • jim_thompson5910
this is a geometric series what's the first term? common ratio?
anonymous
  • anonymous
1/3
jim_thompson5910
  • jim_thompson5910
first term = 1/3, yes
anonymous
  • anonymous
its going up in numerator x2 denomanator x3
jim_thompson5910
  • jim_thompson5910
common ratio = -2/3
anonymous
  • anonymous
okay
anonymous
  • anonymous
so its diverge
anonymous
  • anonymous
right
jim_thompson5910
  • 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
jim_thompson5910
  • jim_thompson5910
Rule: if |r| < 1, then the series converges otherwise, the series diverges
anonymous
  • anonymous
okay so its fixed sum is 1/5
jim_thompson5910
  • jim_thompson5910
if you generated all of the terms and added them up, they would add up to 1/5
anonymous
  • anonymous
okay
anonymous
  • anonymous
so converges are out right
jim_thompson5910
  • jim_thompson5910
what do you mean? I wrote above that the series converges
jim_thompson5910
  • jim_thompson5910
since |r|<1

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