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\[\sum_{n=4}^{15} 4(-3n)\]

\[\sum_{n=1}^{15} 4(-3n)\]

|dw:1439769637735:dw|

|dw:1439769669471:dw|

what are your thoughts?

I was thinking A.

so, why are you thinking A? i need to know your thought process here in order to correct it

Because it looks like an arithmetic sequence and i know that it uses the formula an=a1(n-1)d

No, you're right :)

if its geometric, the common element is multiplied
4(-3) = -12
-12(-3) = 36

yes. I wasn't looking at it like that, but I see your point now.

we also know our index goes from 4 to 15 so that should help eliminate some options as well

but i think you have some of your options displayed incorrectly ...

What do you mean?

|dw:1439770483912:dw|

It means \[\sum_{n=4}^{15}\]

then 4(-3)^n-1

\Large helps magnify the test

??

\Huge is the zoomiest
\[\Huge \sum_{n=start}^{end}\]

ok :) Do you get what I wrote down now?

agreed? if so, what is our solution?

That was how it was written on my math packet so idk

The answer is C?

Ok isn't D saying that then?

the summation of 4 times negative 3 to the n minus 1 power, from n equals 4 to 15.

no, -4 is not 4 on the a_1 part

|dw:1439770907174:dw|

Oh! My mistake, there was no negative 4!

in that case, its fine lol

|dw:1439770953367:dw|
Is this the answer then?

as long as thats not a negative four .. id agree :)

Thanks so much :)

yw

I have another, so I'll just post it in the open question section. K?