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anonymous

  • one year ago

Write the sum using summation notation, assuming the suggested pattern continues. -1 + 2 + 5 + 8 + ... + 44 summation of negative three times n from n equals zero to fifteen summation of the quantity negative one plus three n from n equals zero to fifteen summation of negative three times n from n equals zero to infinity summation of the quantity negative one plus three n from n equals zero to infinity @dan815 @solomonzelman @mathstudent55 @luigi0210 @sammixboo

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  1. Haseeb96
    • one year ago
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    |dw:1436318296947:dw|

  2. anonymous
    • one year ago
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    okay so what do we do next @ha

  3. anonymous
    • one year ago
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    @Haseeb96

  4. misty1212
    • one year ago
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    HI!!

  5. misty1212
    • one year ago
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    they are asking for sigma notation, not a formula for the sum, right?

  6. anonymous
    • one year ago
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    yes @misty1212

  7. Haseeb96
    • one year ago
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    okay okay @misty1212 you should help her I dont know how to solve summatation question and tell us in easy way how to find out the summation

  8. misty1212
    • one year ago
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    \[\sum_{n=-1}^{?}-1+3n\] looks good now we need the upper limit

  9. misty1212
    • one year ago
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    oops that is wrong, i meant \[\sum_{n=0}^{?}-1+3n\] because you are adding 3 each time

  10. Haseeb96
    • one year ago
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    how did you take out this expression (-1+3n) ? @misty1212

  11. anonymous
    • one year ago
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    i think it would be infinity on top

  12. misty1212
    • one year ago
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    to figure out the upper limit, set \[1+3n=44\] solve for \(n\) you should get \(n=15\) making your answer \[\huge\sum_{n=0}^{15}-1+3n\]

  13. misty1212
    • one year ago
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    ok it does say "assuming the pattern continues" so maybe they want \[\sum_{n=0]^{\infty}--+3n\]

  14. misty1212
    • one year ago
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    dang \[\sum_{n=0}^{\infty}-1+3n\]

  15. misty1212
    • one year ago
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    which is kind of silly since that is not a number, but whatever

  16. anonymous
    • one year ago
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    thank you so much @misty1212

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