anonymous
  • anonymous
Use summation notation to write the sum. -4-12-36-...-8748
Mathematics
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anonymous
  • anonymous
Use summation notation to write the sum. -4-12-36-...-8748
Mathematics
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
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anonymous
  • anonymous
\[\sum_{0}^{7} -4*3^n\]
anonymous
  • anonymous
that is correct
anonymous
  • anonymous
i think

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anonymous
  • anonymous
Ok but could you please explain how you got that because this is a practice problem before I take my test...
anonymous
  • anonymous
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
anonymous
  • anonymous
Ok now I have a choice of answers and yours does not match up exactly to any of them but it is close....
anonymous
  • anonymous
So then you just need to find where to stop by finding which power of 3 multiplied by -4 will yield 8748. I used\[\log_{3} 2187 = 7\]
anonymous
  • anonymous
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.
anonymous
  • anonymous
\[\sum_{1}^{8}-4\times3^{n-1}\] do any of your answers correspond to this instead? same thing, just diff notation
anonymous
  • anonymous
Yes One does... How did you come up with that?
anonymous
  • anonymous
you get that if you take -4 as the 1st term, rather than the 0th term
anonymous
  • anonymous
oh ok that makes sense.. Thanks!
anonymous
  • anonymous
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)
anonymous
  • anonymous
ohhhh ok I gotcha now! Thanks a lot!
anonymous
  • anonymous
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.
anonymous
  • anonymous
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
anonymous
  • anonymous
cool thanks ya'll!
anonymous
  • anonymous
What about...Find the sum of the infinite series. \[\sum_{i=1}^{\infty}2(-1/4)^i\]

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