anonymous
  • anonymous
Sigma Notation
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
\[\sum_{n=4}^{15} 4(-3n)\]
anonymous
  • anonymous
\[\sum_{n=1}^{15} 4(-3n)\]
anonymous
  • anonymous
|dw:1439769637735:dw|

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anonymous
  • anonymous
|dw:1439769669471:dw|
anonymous
  • anonymous
@amistre64
amistre64
  • amistre64
what are your thoughts?
anonymous
  • anonymous
I was thinking A.
amistre64
  • amistre64
so, why are you thinking A? i need to know your thought process here in order to correct it
anonymous
  • anonymous
Because it looks like an arithmetic sequence and i know that it uses the formula an=a1(n-1)d
amistre64
  • amistre64
hmm, if it was arithmetic, then the difference between terms is constant 4 + (-16) = -12 -12 + (-16) .. is not 36, is it?
anonymous
  • anonymous
No, you're right :)
amistre64
  • amistre64
if its geometric, the common element is multiplied 4(-3) = -12 -12(-3) = 36
anonymous
  • anonymous
yes. I wasn't looking at it like that, but I see your point now.
amistre64
  • amistre64
we also know our index goes from 4 to 15 so that should help eliminate some options as well
amistre64
  • amistre64
but i think you have some of your options displayed incorrectly ...
anonymous
  • anonymous
What do you mean?
amistre64
  • amistre64
|dw:1439770483912:dw|
anonymous
  • anonymous
It means \[\sum_{n=4}^{15}\]
anonymous
  • anonymous
then 4(-3)^n-1
amistre64
  • amistre64
\Large helps magnify the test
anonymous
  • anonymous
??
amistre64
  • amistre64
\Huge is the zoomiest \[\Huge \sum_{n=start}^{end}\]
anonymous
  • anonymous
ok :) Do you get what I wrote down now?
amistre64
  • 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
  • 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
  • amistre64
agreed? if so, what is our solution?
anonymous
  • anonymous
That was how it was written on my math packet so idk
anonymous
  • anonymous
The answer is C?
amistre64
  • 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
  • anonymous
Ok isn't D saying that then?
anonymous
  • anonymous
the summation of 4 times negative 3 to the n minus 1 power, from n equals 4 to 15.
amistre64
  • amistre64
no, -4 is not 4 on the a_1 part
amistre64
  • amistre64
|dw:1439770907174:dw|
anonymous
  • anonymous
Oh! My mistake, there was no negative 4!
amistre64
  • amistre64
in that case, its fine lol
anonymous
  • anonymous
|dw:1439770953367:dw| Is this the answer then?
amistre64
  • amistre64
as long as thats not a negative four .. id agree :)
anonymous
  • anonymous
Thanks so much :)
amistre64
  • amistre64
yw
anonymous
  • anonymous
I have another, so I'll just post it in the open question section. K?
amistre64
  • amistre64
k

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