anonymous
  • anonymous
The wildflowers at a national park have been decreasing in numbers. There were 300 wildflowers in the first year that the park started tracking them. Since then, there have been one fourth as many new flowers each year. Create the sigma notation showing the infinite growth of the wildflowers and find the sum, if possible.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
|dw:1440074880297:dw|
anonymous
  • anonymous
\[\sum_{i=1}^{\infty}(\frac{ 1 }{4}^i ; 400\]
anonymous
  • anonymous
\[\sum_{i=1}^{\infty}1200(\frac{ 1 }{ 4 })^i\] no sum bc it's diveregent

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anonymous
  • anonymous
|dw:1440075264298:dw|
anonymous
  • anonymous
|dw:1440075318000:dw|
phi
  • phi
you want to start the summation at i=0 with the 1200 that makes the first term 1200 in other words \[ \sum_{i=0}^\infty 1200\left({ \frac{1}{4}}\right)^i = 1200 + \frac{1200}{4} + \frac{1200}{16}+...\]
phi
  • phi
remember (1/4)^0 is 1 (anything to the zero power is 1 ... except 0^0 which is not defined) the other idea you should learn is a fraction smaller than 1 gets smaller when you multiply it by itself 1/4 *1/4 = 1/16 and 1/16 is smaller than 1/4 as we keep multiply, we get numbers like 1/1024 (about 0.001) and eventually, we get a *very* tiny number, which we can treat as "practically zero" and that makes the series convergent (because at some point we are not adding any more terms (they are *so tiny* they don't make the sum bigger)
anonymous
  • anonymous
Ok, so since it's convergent would the exponent be i-1?
anonymous
  • anonymous
I got C :)??
anonymous
  • anonymous
@ganeshie8
phi
  • phi
you can write the series so i starts at 0 (see up above) once you do that, find 1/(1-r) then multiply by 1200
phi
  • phi
1/(1-1/4) = 1/(3/4) = 4/3 and the sum is 1200*4/3 = 400*4= 1600
anonymous
  • anonymous
Thanks :)!!
phi
  • phi
of course, if you start the summation at i=1 then you make the exponent i-1
anonymous
  • anonymous
Yup, then you can just narrow it down from there:)

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