Sigma Notation

- anonymous

Sigma Notation

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- schrodinger

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- anonymous

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

- anonymous

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

- anonymous

|dw:1439769637735:dw|

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## More answers

- anonymous

|dw:1439769669471:dw|

- anonymous

@amistre64

- amistre64

what are your thoughts?

- anonymous

I was thinking A.

- amistre64

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

- anonymous

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

- amistre64

hmm, if it was arithmetic, then the difference between terms is constant
4 + (-16) = -12
-12 + (-16) .. is not 36, is it?

- anonymous

No, you're right :)

- amistre64

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

- anonymous

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

- amistre64

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

- amistre64

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

- anonymous

What do you mean?

- amistre64

|dw:1439770483912:dw|

- anonymous

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

- anonymous

then 4(-3)^n-1

- amistre64

\Large helps magnify the test

- anonymous

??

- amistre64

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

- anonymous

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

- amistre64

we want to start at 4 and end at 15 right? ... but the only real option we have i believe you misprinted ..
|dw:1439770648146:dw|

- amistre64

with a geometric sequence, the series is represented as an exponent on the common ratio
\[\Huge \sum_{n=start}^{end}a_1(r)^{n-1}\]

- amistre64

agreed? if so, what is our solution?

- anonymous

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

- anonymous

The answer is C?

- amistre64

hmm, then somewhere there is a misprint. C is 'closest' we have to the right format; but we would want it to start at 4, not 1
\[\Huge \sum_{n=\color{red} 4}^{15}4(-3)^{n-1}\]

- anonymous

Ok isn't D saying that then?

- anonymous

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

- amistre64

no, -4 is not 4 on the a_1 part

- amistre64

|dw:1439770907174:dw|

- anonymous

Oh! My mistake, there was no negative 4!

- amistre64

in that case, its fine lol

- anonymous

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

- amistre64

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

- anonymous

Thanks so much :)

- amistre64

yw

- anonymous

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

- amistre64

k

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