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anonymous

  • one year ago

Can someone please help me with this Sigma Notation question?????

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  1. anonymous
    • one year ago
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  2. anonymous
    • one year ago
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    sure i just finished a test on it xD

  3. anonymous
    • one year ago
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    wit u need help bout?

  4. SolomonZelman
    • one year ago
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    It is a geometric sequence, I hope that is not necessary to be mentioned and is well known to all..... right?

  5. anonymous
    • one year ago
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    lol if u take the class its a basic

  6. SolomonZelman
    • one year ago
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    \(\large\color{black}{ \displaystyle \sum_{ n=2 }^{ 10 } ~25(0.3)^{n+1}}\) (just putting it for vision, not to look at that tab again...

  7. SolomonZelman
    • one year ago
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    \(\large\color{black}{ \displaystyle \sum_{ n=2 }^{ 10 } ~25(0.3)^{n+1}}\) 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

  8. SolomonZelman
    • one year ago
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    No, you are starting from n=2, Icedragon50

  9. anonymous
    • one year ago
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    oooohhh

  10. SolomonZelman
    • one year ago
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    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 \(25(0.3)^{n+1}\).

  11. SolomonZelman
    • one year ago
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    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 ?

  12. anonymous
    • one year ago
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    r=0.3

  13. anonymous
    • one year ago
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    Isn't it 0.3

  14. SolomonZelman
    • one year ago
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    yes, r=0.3

  15. SolomonZelman
    • one year ago
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    So, 9 terms, r=0.3 Now we need our first term, and in that case \(a_2\)

  16. anonymous
    • one year ago
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    a1=25(0.3)^2+1

  17. SolomonZelman
    • one year ago
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    you need \(a_\color{red}{2}\)

  18. SolomonZelman
    • one year ago
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    that is what we start from and not ace of 1.

  19. SolomonZelman
    • one year ago
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    \(\large\color{black}{ \displaystyle a_2=25(0.3)^{n+2}=25(0.3)^3=25(27/1000)=27/40}\)

  20. SolomonZelman
    • one year ago
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    \(\large\color{black}{ \displaystyle \sum_{ n=2 }^{ 10 } ~25(0.3)^{n+1}=\frac{(27/40)\cdot (1-0.3^9)}{1-0.3}}\)

  21. SolomonZelman
    • one year ago
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    same way you can do the first ten terms w/ \(a_1\), but to then subtract \(a_1\) from the sum.

  22. anonymous
    • one year ago
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    he's probably just gonna look at the answer since he's not responding so bai bai lol

  23. anonymous
    • one year ago
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    My computer actually died and I fell asleep, but ok.

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