anonymous
  • anonymous
Can someone please help me with this Sigma Notation question?????
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
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anonymous
  • anonymous
sure i just finished a test on it xD
anonymous
  • anonymous
wit u need help bout?

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SolomonZelman
  • SolomonZelman
It is a geometric sequence, I hope that is not necessary to be mentioned and is well known to all..... right?
anonymous
  • anonymous
lol if u take the class its a basic
SolomonZelman
  • SolomonZelman
\(\large\color{black}{ \displaystyle \sum_{ n=2 }^{ 10 } ~25(0.3)^{n+1}}\) (just putting it for vision, not to look at that tab again...
SolomonZelman
  • SolomonZelman
\(\large\color{black}{ \displaystyle \sum_{ n=2 }^{ 10 } ~25(0.3)^{n+1}}\) you need to know: 1) the common ratio. 2) the number of terms that you are adding 3) the (value of the) first term that you start from
SolomonZelman
  • SolomonZelman
No, you are starting from n=2, Icedragon50
anonymous
  • anonymous
oooohhh
SolomonZelman
  • SolomonZelman
The total number of terms is 10-2=8. (But this is not including the \(a_2\) and so, with the \(a_2\) you got 9 terms. Your first term that you start adding is \(a_2\) (then \(a_3\), \(a_4\), \(a_5\) and so on... till the \(a_{10}\)). And to find this first term plug in n=2 into \(25(0.3)^{n+1}\).
SolomonZelman
  • SolomonZelman
The common ratio is what you multiply by. You can figure the common ratio (r) simply by writing out a couple of terms. \(\large\color{black}{ \displaystyle a_2=25(0.3)^{2+1}=25(0.3)^3}\) \(\large\color{black}{ \displaystyle a_2=25(0.3)^{2+2}=25(0.3)^4}\) \(\large\color{black}{ \displaystyle a_2=25(0.3)^{2+3}=25(0.3)^5}\) \(\large\color{black}{ \displaystyle a_2=25(0.3)^{2+4}=25(0.3)^6}\) \(\large\color{black}{ \displaystyle a_2=25(0.3)^{2+5}=25(0.3)^7}\) And so on... can you give me the common ratio ?
anonymous
  • anonymous
r=0.3
anonymous
  • anonymous
Isn't it 0.3
SolomonZelman
  • SolomonZelman
yes, r=0.3
SolomonZelman
  • SolomonZelman
So, 9 terms, r=0.3 Now we need our first term, and in that case \(a_2\)
anonymous
  • anonymous
a1=25(0.3)^2+1
SolomonZelman
  • SolomonZelman
you need \(a_\color{red}{2}\)
SolomonZelman
  • SolomonZelman
that is what we start from and not ace of 1.
SolomonZelman
  • SolomonZelman
\(\large\color{black}{ \displaystyle a_2=25(0.3)^{n+2}=25(0.3)^3=25(27/1000)=27/40}\)
SolomonZelman
  • SolomonZelman
\(\large\color{black}{ \displaystyle \sum_{ n=2 }^{ 10 } ~25(0.3)^{n+1}=\frac{(27/40)\cdot (1-0.3^9)}{1-0.3}}\)
SolomonZelman
  • SolomonZelman
same way you can do the first ten terms w/ \(a_1\), but to then subtract \(a_1\) from the sum.
anonymous
  • anonymous
he's probably just gonna look at the answer since he's not responding so bai bai lol
anonymous
  • anonymous
My computer actually died and I fell asleep, but ok.

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