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anonymous

  • one year ago

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.

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  1. anonymous
    • one year ago
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    |dw:1440074880297:dw|

  2. anonymous
    • one year ago
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    \[\sum_{i=1}^{\infty}(\frac{ 1 }{4}^i ; 400\]

  3. anonymous
    • one year ago
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    \[\sum_{i=1}^{\infty}1200(\frac{ 1 }{ 4 })^i\] no sum bc it's diveregent

  4. anonymous
    • one year ago
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    |dw:1440075264298:dw|

  5. anonymous
    • one year ago
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    |dw:1440075318000:dw|

  6. phi
    • one year ago
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    you want to start the summation at i=0 with the 1200 that makes the first term 1200 in other words \[ \sum_{i=0}^\infty 1200\left({ \frac{1}{4}}\right)^i = 1200 + \frac{1200}{4} + \frac{1200}{16}+...\]

  7. phi
    • one year ago
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    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)

  8. anonymous
    • one year ago
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    Ok, so since it's convergent would the exponent be i-1?

  9. anonymous
    • one year ago
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    I got C :)??

  10. anonymous
    • one year ago
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    @ganeshie8

  11. phi
    • one year ago
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    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

  12. phi
    • one year ago
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    1/(1-1/4) = 1/(3/4) = 4/3 and the sum is 1200*4/3 = 400*4= 1600

  13. anonymous
    • one year ago
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    Thanks :)!!

  14. phi
    • one year ago
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    of course, if you start the summation at i=1 then you make the exponent i-1

  15. anonymous
    • one year ago
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    Yup, then you can just narrow it down from there:)

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