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Babynini

  • one year ago

Find the partial sum S_n of the geometric sequence that satisfies the given conditions. 10 over sigma k=0 beneath 3(1/2)^k on the side.

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  1. Babynini
    • one year ago
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    @jim_thompson5910 if you're available!

  2. Empty
    • one year ago
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    \[\sum_{k=0}^{10} 3 \left(\frac{1}{2} \right)^k\] Right? Any ideas how to simplify this? Do you know the geometric series formula or know how to derive it? here's the code by the way: ``` \[\sum_{k=0}^{10} 3 \left(\frac{1}{2} \right)^k\] ```

  3. Babynini
    • one year ago
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    yeah that's it :)

  4. Babynini
    • one year ago
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    er they've had me using the formula s_n=a([1-r^n]/[1-r]) for the sum.

  5. Babynini
    • one year ago
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    do you mean if I know how to write that out in terms?

  6. Empty
    • one year ago
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    Ok yeah that formula works, do you know what "a" and "r" and "n" are? Can you fill in the question marks? \[s_n=a \frac{1-r^n}{1-r}=\sum_{k=0}^{?} ?\] ``` \[s_n=a \frac{1-r^n}{1-r}=\sum_{k=0}^{?} ?\] ```

  7. Babynini
    • one year ago
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    oh oh I see. \[s_n=a \frac{1-r^n}{1-r}=\sum_{k=0}^{n} ?\] like that?

  8. Babynini
    • one year ago
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    \[s_n=a \frac{1-r^n}{1-r}=\sum_{k=0}^{n} a(r)\] is that correct? o.o

  9. Empty
    • one year ago
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    Almost not quite, since if you plug in n=1 you would get what? \[a \frac{1-r^1}{1-r} = 1\] for the left side and \[\sum_{k=0}^{1} a(r) = ar+ar=2ar\] on the right side. I'll give you a hint, try this, add a power of k in there: (for this example I picked 2 on the top to show how I calculate it on the right, but it also has a geometric series that I haven't shown you that it's equal to.) \[\sum_{k=0}^{2} a*r^k= ar^0+ar^1+ar^2\] But the real value on top will be n-1, not n. \[a\frac{1-r^n}{1-r}=\sum_{k=0}^{n-1} a*r^k\]

  10. Babynini
    • one year ago
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    ah, sorry. Pc got messed up but i'm back now!

  11. Babynini
    • one year ago
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    oh hrm.

  12. Babynini
    • one year ago
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    so what do we do for the sum formula?

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