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anonymous
 one year ago
Can someone please help me with this Sigma Notation question?????
anonymous
 one year ago
Can someone please help me with this Sigma Notation question?????

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0sure i just finished a test on it xD

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0wit u need help bout?

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1It is a geometric sequence, I hope that is not necessary to be mentioned and is well known to all..... right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0lol if u take the class its a basic

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1\(\large\color{black}{ \displaystyle \sum_{ n=2 }^{ 10 } ~25(0.3)^{n+1}}\) (just putting it for vision, not to look at that tab again...

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1\(\large\color{black}{ \displaystyle \sum_{ n=2 }^{ 10 } ~25(0.3)^{n+1}}\) 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
 one year ago
Best ResponseYou've already chosen the best response.1No, you are starting from n=2, Icedragon50

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1The total number of terms is 102=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
 one year ago
Best ResponseYou've already chosen the best response.1The 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 ?

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1So, 9 terms, r=0.3 Now we need our first term, and in that case \(a_2\)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1you need \(a_\color{red}{2}\)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1that is what we start from and not ace of 1.

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1\(\large\color{black}{ \displaystyle a_2=25(0.3)^{n+2}=25(0.3)^3=25(27/1000)=27/40}\)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1\(\large\color{black}{ \displaystyle \sum_{ n=2 }^{ 10 } ~25(0.3)^{n+1}=\frac{(27/40)\cdot (10.3^9)}{10.3}}\)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1same way you can do the first ten terms w/ \(a_1\), but to then subtract \(a_1\) from the sum.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0he's probably just gonna look at the answer since he's not responding so bai bai lol

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0My computer actually died and I fell asleep, but ok.
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