Question

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

Answer 1

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

Answer 2

that is correct

Answer 3

i think

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

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.

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

oh ok that makes sense.. Thanks!

Answer 13

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)

Answer 14

ohhhh ok I gotcha now! Thanks a lot!

Answer 15

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.

Answer 16

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

Answer 17

cool thanks ya'll!

Answer 18

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