anonymous
  • anonymous
Evaluate this series.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
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anonymous
  • anonymous
@peachpi
anonymous
  • anonymous
That's small enough where you can find each term and add them together. Plug in 1, 2, 3, 4, 5, and 6 for k then add the results

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Astrophysics
  • Astrophysics
They should say "expand the series"
Astrophysics
  • Astrophysics
expand and evaluate actually
Loser66
  • Loser66
@peachpi what if we calculate by expand the sum? like \[\sum_{k=1}^6 2k^2 +\sum_{k=1}^6 k-\sum_{k=1}^6 5\] ??
anonymous
  • anonymous
I know you can use the arithmetic series formula for the second one and the third one is 30. I don't know if there's a handy way to do the first sum
Loser66
  • Loser66
we have all formula for them \[\sum_{k=1}^6 2k^2=2\sum_{k=1}^6 k^2=2\dfrac{6*(6+1)*(2*6+1)}{6}= 210\]
Loser66
  • Loser66
but what I don't know is I can't get 173 which is the right answer if we just "plug in" ha!!
Loser66
  • Loser66
for the middle term, we have \(\dfrac{6(6+1)}{2}=21\) the last term is just -5 hence 210+21 -5 \(\neq 173\) ha!!
Loser66
  • Loser66
@dan815 what is wrong with my logic???
Loser66
  • Loser66
@peachpi you said the last one is -30?
anonymous
  • anonymous
yes -30 because you're subtracting 5 six times
Loser66
  • Loser66
But if it is -30, we can't get 173 though!! 210+21 -30 = 201
Loser66
  • Loser66
@jim_thompson5910 Hello Mr. Jim, please, help me figure out what is wrong.
anonymous
  • anonymous
I got 182 for the first one
jim_thompson5910
  • jim_thompson5910
yeah agreed. The sum of the k^2 terms is 91 which doubles to 182
Loser66
  • Loser66
@peachy Thanks a lot. My mistake, :) I calculate 2n+1 = 15, not 13. @jim_thompson5910 Thanks for quickly respond.
anonymous
  • anonymous
Oh, I got it thank you guys! I have another question.
anonymous
  • anonymous
The population of blue-green algae in a certain lake is increasing at a rate of 33% per hour. The count was at 5,000 cells per milliliter at noon on August 1. By what date and time will the algae count get to the "high alert" level of 50,000 cells per milliliter? Assume that the algae continues to increase at the same rate.
jim_thompson5910
  • jim_thompson5910
\[\Large F = P(1+r)^t\] \[\Large 50,000 = 5,000(1+0.33)^t\] solve for t t = number of hours (t = 0 corresponds to noon on Aug 1st)
anonymous
  • anonymous
Oh, I understand it now! Thank you!
jim_thompson5910
  • jim_thompson5910
no problem

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