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anonymous

  • 5 years ago

Use summation notation to write the sum. -4-12-36-...-8748

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  1. anonymous
    • 5 years ago
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    \[\sum_{0}^{7} -4*3^n\]

  2. anonymous
    • 5 years ago
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    that is correct

  3. anonymous
    • 5 years ago
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    i think

  4. anonymous
    • 5 years ago
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    Ok but could you please explain how you got that because this is a practice problem before I take my test...

  5. anonymous
    • 5 years ago
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    Well, you can see that each term is 3 times the previous one. So your sequence is of powers of 3. The first term is -4 though which must corrispond to the 0th power of 3 since there are no 3's in its factored form

  6. anonymous
    • 5 years ago
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    Ok now I have a choice of answers and yours does not match up exactly to any of them but it is close....

  7. anonymous
    • 5 years ago
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    So then you just need to find where to stop by finding which power of 3 multiplied by -4 will yield 8748. I used\[\log_{3} 2187 = 7\]

  8. anonymous
    • 5 years ago
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    That is the correct answer. You can find the correct power by simply counting and multiplying by 3 until you get to 2187, but it'll be the same. If that's not one of the possible answers then something is wrong with the way you've written the question here.

  9. anonymous
    • 5 years ago
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    \[\sum_{1}^{8}-4\times3^{n-1}\] do any of your answers correspond to this instead? same thing, just diff notation

  10. anonymous
    • 5 years ago
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    Yes One does... How did you come up with that?

  11. anonymous
    • 5 years ago
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    you get that if you take -4 as the 1st term, rather than the 0th term

  12. anonymous
    • 5 years ago
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    oh ok that makes sense.. Thanks!

  13. anonymous
    • 5 years ago
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    Yeah, changing your index to start at 1 just means that each of your n's will be 1 more than each of mine, Therefore you will stop at 8 and multiply 4 by 3^(n-1)

  14. anonymous
    • 5 years ago
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    ohhhh ok I gotcha now! Thanks a lot!

  15. anonymous
    • 5 years ago
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    You can also write it as going from 2 to 9 and use n-2 or 3-10 and use n-3, etc. They will still be the same sum because they will have identical terms.

  16. anonymous
    • 5 years ago
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    wouldn't it make more sense to take it -4 as the first term though? because having a '0th' term implies that there is nothing there...but yeah, either way you'll get the same answer

  17. anonymous
    • 5 years ago
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    cool thanks ya'll!

  18. anonymous
    • 5 years ago
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    What about...Find the sum of the infinite series. \[\sum_{i=1}^{\infty}2(-1/4)^i\]

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