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anonymous
 one year ago
Evaluate this series.
anonymous
 one year ago
Evaluate this series.

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0That'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

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1They should say "expand the series"

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1expand and evaluate actually

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0@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
 one year ago
Best ResponseYou've already chosen the best response.0I 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
 one year ago
Best ResponseYou've already chosen the best response.0we 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
 one year ago
Best ResponseYou've already chosen the best response.0but what I don't know is I can't get 173 which is the right answer if we just "plug in" ha!!

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0for 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
 one year ago
Best ResponseYou've already chosen the best response.0@dan815 what is wrong with my logic???

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0@peachpi you said the last one is 30?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes 30 because you're subtracting 5 six times

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0But if it is 30, we can't get 173 though!! 210+21 30 = 201

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0@jim_thompson5910 Hello Mr. Jim, please, help me figure out what is wrong.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I got 182 for the first one

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1yeah agreed. The sum of the k^2 terms is 91 which doubles to 182

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0@peachy Thanks a lot. My mistake, :) I calculate 2n+1 = 15, not 13. @jim_thompson5910 Thanks for quickly respond.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh, I got it thank you guys! I have another question.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The population of bluegreen 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
 one year ago
Best ResponseYou've already chosen the best response.1\[\Large F = P(1+r)^t\] \[\Large 50,000 = 5,000(1+0.33)^t\] solve for t t = number of hours (t = 0 corresponds to noon on Aug 1st)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh, I understand it now! Thank you!
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