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anonymous

  • one year ago

Evaluate this series.

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  1. anonymous
    • one year ago
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  2. anonymous
    • one year ago
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    @peachpi

  3. anonymous
    • one year ago
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    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

  4. Astrophysics
    • one year ago
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    They should say "expand the series"

  5. Astrophysics
    • one year ago
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    expand and evaluate actually

  6. Loser66
    • one year ago
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    @peachpi what if we calculate by expand the sum? like \[\sum_{k=1}^6 2k^2 +\sum_{k=1}^6 k-\sum_{k=1}^6 5\] ??

  7. anonymous
    • one year ago
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    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

  8. Loser66
    • one year ago
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    we have all formula for them \[\sum_{k=1}^6 2k^2=2\sum_{k=1}^6 k^2=2\dfrac{6*(6+1)*(2*6+1)}{6}= 210\]

  9. Loser66
    • one year ago
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    but what I don't know is I can't get 173 which is the right answer if we just "plug in" ha!!

  10. Loser66
    • one year ago
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    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!!

  11. Loser66
    • one year ago
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    @dan815 what is wrong with my logic???

  12. Loser66
    • one year ago
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    @peachpi you said the last one is -30?

  13. anonymous
    • one year ago
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    yes -30 because you're subtracting 5 six times

  14. Loser66
    • one year ago
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    But if it is -30, we can't get 173 though!! 210+21 -30 = 201

  15. Loser66
    • one year ago
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    @jim_thompson5910 Hello Mr. Jim, please, help me figure out what is wrong.

  16. anonymous
    • one year ago
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    I got 182 for the first one

  17. jim_thompson5910
    • one year ago
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    yeah agreed. The sum of the k^2 terms is 91 which doubles to 182

  18. Loser66
    • one year ago
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    @peachy Thanks a lot. My mistake, :) I calculate 2n+1 = 15, not 13. @jim_thompson5910 Thanks for quickly respond.

  19. anonymous
    • one year ago
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    Oh, I got it thank you guys! I have another question.

  20. anonymous
    • one year ago
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    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.

  21. jim_thompson5910
    • one year ago
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    \[\Large F = P(1+r)^t\] \[\Large 50,000 = 5,000(1+0.33)^t\] solve for t t = number of hours (t = 0 corresponds to noon on Aug 1st)

  22. anonymous
    • one year ago
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    Oh, I understand it now! Thank you!

  23. jim_thompson5910
    • one year ago
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    no problem

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