anonymous
  • anonymous
The population of a local species of dragonfly can be found using an infinite geometric series where a1 = 42 and the common ratio is three fourths. Write the sum in sigma notation and calculate the sum that will be the upper limit of this population.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
@jim_thompson5910
jim_thompson5910
  • jim_thompson5910
what do you have so far?
anonymous
  • anonymous
i have|dw:1442973187600:dw|

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jim_thompson5910
  • jim_thompson5910
you're very close
anonymous
  • anonymous
oh and the 42(3/4)^i-1?
jim_thompson5910
  • jim_thompson5910
it should be \[\Large \sum_{i=1}^{\infty} 42\left(\frac{3}{4}\right)^{i-1}\]
anonymous
  • anonymous
Yes! That's what I meant. Thank-you :)
jim_thompson5910
  • jim_thompson5910
now `calculate the sum that will be the upper limit of this population.`
anonymous
  • anonymous
Hmmm...it wouldn't be diveregent right?
anonymous
  • anonymous
divergent*
jim_thompson5910
  • jim_thompson5910
you see how a = 42 and r = 3/4 right?
anonymous
  • anonymous
yes
anonymous
  • anonymous
ok! So we are plugging it into the geometric series formula!
jim_thompson5910
  • jim_thompson5910
yeah since |r| < 1 is true, we can use S = a/(1-r)
anonymous
  • anonymous
ok. So I got 168?
jim_thompson5910
  • jim_thompson5910
me too
anonymous
  • anonymous
Can you help me on another?
jim_thompson5910
  • jim_thompson5910
sure
anonymous
  • anonymous
What is the sum of the arithmetic sequence 8, 14, 20 …, if there are 24 terms?
jim_thompson5910
  • jim_thompson5910
first term = ?? common difference = ??
anonymous
  • anonymous
A1=8 C.D.=-6
jim_thompson5910
  • jim_thompson5910
C.D is +6 actually
jim_thompson5910
  • jim_thompson5910
we add 6 to each term to get the next one
anonymous
  • anonymous
oh! Sorry :)
jim_thompson5910
  • jim_thompson5910
now use the formula \[\Large S_n = \frac{n(2a+d(n-1))}{2}\] n = number of terms added up a = first term d = common difference
anonymous
  • anonymous
So far, I have s24=24(16+6n-1/2)
jim_thompson5910
  • jim_thompson5910
n = 24, so replace the n in n-1 with 24
anonymous
  • anonymous
oh! Thanks
jim_thompson5910
  • jim_thompson5910
\[\Large S_n = \frac{n(2a+d(n-1))}{2}\] \[\Large S_{24} = \frac{24(2*8+6(24-1))}{2}\] \[\Large S_{24} = ???\]
anonymous
  • anonymous
Ok! Hold on lol
anonymous
  • anonymous
1848 :)
jim_thompson5910
  • jim_thompson5910
good

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