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anonymous
 5 years ago
Use summation notation to write the sum.
41236...8748
anonymous
 5 years ago
Use summation notation to write the sum. 41236...8748

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\sum_{0}^{7} 4*3^n\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ok but could you please explain how you got that because this is a practice problem before I take my test...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well, 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
 5 years ago
Best ResponseYou've already chosen the best response.0Ok now I have a choice of answers and yours does not match up exactly to any of them but it is close....

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So 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
 5 years ago
Best ResponseYou've already chosen the best response.0That 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
 5 years ago
Best ResponseYou've already chosen the best response.0\[\sum_{1}^{8}4\times3^{n1}\] do any of your answers correspond to this instead? same thing, just diff notation

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yes One does... How did you come up with that?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you get that if you take 4 as the 1st term, rather than the 0th term

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh ok that makes sense.. Thanks!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yeah, 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^(n1)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ohhhh ok I gotcha now! Thanks a lot!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You can also write it as going from 2 to 9 and use n2 or 310 and use n3, etc. They will still be the same sum because they will have identical terms.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0wouldn'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
 5 years ago
Best ResponseYou've already chosen the best response.0What about...Find the sum of the infinite series. \[\sum_{i=1}^{\infty}2(1/4)^i\]
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