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anonymous

  • one year ago

Sigma Notation

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  1. anonymous
    • one year ago
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    \[\sum_{n=4}^{15} 4(-3n)\]

  2. anonymous
    • one year ago
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    \[\sum_{n=1}^{15} 4(-3n)\]

  3. anonymous
    • one year ago
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    |dw:1439769637735:dw|

  4. anonymous
    • one year ago
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    |dw:1439769669471:dw|

  5. anonymous
    • one year ago
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    @amistre64

  6. amistre64
    • one year ago
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    what are your thoughts?

  7. anonymous
    • one year ago
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    I was thinking A.

  8. amistre64
    • one year ago
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    so, why are you thinking A? i need to know your thought process here in order to correct it

  9. anonymous
    • one year ago
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    Because it looks like an arithmetic sequence and i know that it uses the formula an=a1(n-1)d

  10. amistre64
    • one year ago
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    hmm, if it was arithmetic, then the difference between terms is constant 4 + (-16) = -12 -12 + (-16) .. is not 36, is it?

  11. anonymous
    • one year ago
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    No, you're right :)

  12. amistre64
    • one year ago
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    if its geometric, the common element is multiplied 4(-3) = -12 -12(-3) = 36

  13. anonymous
    • one year ago
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    yes. I wasn't looking at it like that, but I see your point now.

  14. amistre64
    • one year ago
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    we also know our index goes from 4 to 15 so that should help eliminate some options as well

  15. amistre64
    • one year ago
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    but i think you have some of your options displayed incorrectly ...

  16. anonymous
    • one year ago
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    What do you mean?

  17. amistre64
    • one year ago
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    |dw:1439770483912:dw|

  18. anonymous
    • one year ago
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    It means \[\sum_{n=4}^{15}\]

  19. anonymous
    • one year ago
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    then 4(-3)^n-1

  20. amistre64
    • one year ago
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    \Large helps magnify the test

  21. anonymous
    • one year ago
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    ??

  22. amistre64
    • one year ago
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    \Huge is the zoomiest \[\Huge \sum_{n=start}^{end}\]

  23. anonymous
    • one year ago
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    ok :) Do you get what I wrote down now?

  24. amistre64
    • one year ago
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    we want to start at 4 and end at 15 right? ... but the only real option we have i believe you misprinted .. |dw:1439770648146:dw|

  25. amistre64
    • one year ago
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    with a geometric sequence, the series is represented as an exponent on the common ratio \[\Huge \sum_{n=start}^{end}a_1(r)^{n-1}\]

  26. amistre64
    • one year ago
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    agreed? if so, what is our solution?

  27. anonymous
    • one year ago
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    That was how it was written on my math packet so idk

  28. anonymous
    • one year ago
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    The answer is C?

  29. amistre64
    • one year ago
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    hmm, then somewhere there is a misprint. C is 'closest' we have to the right format; but we would want it to start at 4, not 1 \[\Huge \sum_{n=\color{red} 4}^{15}4(-3)^{n-1}\]

  30. anonymous
    • one year ago
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    Ok isn't D saying that then?

  31. anonymous
    • one year ago
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    the summation of 4 times negative 3 to the n minus 1 power, from n equals 4 to 15.

  32. amistre64
    • one year ago
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    no, -4 is not 4 on the a_1 part

  33. amistre64
    • one year ago
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    |dw:1439770907174:dw|

  34. anonymous
    • one year ago
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    Oh! My mistake, there was no negative 4!

  35. amistre64
    • one year ago
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    in that case, its fine lol

  36. anonymous
    • one year ago
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    |dw:1439770953367:dw| Is this the answer then?

  37. amistre64
    • one year ago
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    as long as thats not a negative four .. id agree :)

  38. anonymous
    • one year ago
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    Thanks so much :)

  39. amistre64
    • one year ago
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    yw

  40. anonymous
    • one year ago
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    I have another, so I'll just post it in the open question section. K?

  41. amistre64
    • one year ago
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    k

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