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

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- anonymous

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

- chestercat

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- anonymous

\[\sum_{0}^{7} -4*3^n\]

- anonymous

that is correct

- anonymous

i think

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## More answers

- anonymous

Ok but could you please explain how you got that because this is a practice problem before I take my test...

- 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

Ok now I have a choice of answers and yours does not match up exactly to any of them but it is close....

- 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

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

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

- anonymous

Yes One does... How did you come up with that?

- anonymous

you get that if you take -4 as the 1st term, rather than the 0th term

- anonymous

oh ok that makes sense.. Thanks!

- 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

ohhhh ok I gotcha now! Thanks a lot!

- 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

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

cool thanks ya'll!

- anonymous

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

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