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anonymous

  • 5 years ago

For the geometric series 1 + 3 + 9 + 27 + . . . , find the sum of the first 10 terms. does anyone know how to solve this?

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  1. anonymous
    • 5 years ago
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    each one is *3 so 1+3+9+27+81+243+729+2187+6561+19683 = 29524. the correct way to do this is using factorials (the exclamation point symbol) factorial of 3 over a range of 10

  2. anonymous
    • 5 years ago
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    Loss, that isn't help ... "sum it by writing out ever terms and adding them ' ... good job.

  3. anonymous
    • 5 years ago
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    I'll post the correct method:

  4. anonymous
    • 5 years ago
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    ok thanks

  5. anonymous
    • 5 years ago
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    Call the geometric series, Sn: \[Sn = a + ar + ar^2 + ... + ar^{n-1}\] multiplying by r: \[rSn = ar + ar^2 + ar^3 + ... + ar^{n}\] So we can say: \[Sn - rSn = a - ar^n \implies Sn(1-r) = a(1-r^n)\] \[ \implies Sn = \frac{a(1-r^n)}{1-r}\] This is the general sum or a geometric series with starting term a and ratio r Can you see what your a and r are, and hence solve?

  6. anonymous
    • 5 years ago
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    * and n terms

  7. anonymous
    • 5 years ago
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    Oh, OK:( a = 1, r = 3, n = 2 \[\implies Sn = \frac{1-3^{10}}{1-3} = 29524\] Oh, wow, it's the same as above.. except now I can solve any general geometric series. Taking the above further: \[Sn = \frac{a(1-r^n)}{1-r}\] If |r| < 1, \[r^n \rightarrow 0\ \text{as n } \rightarrow \infty\] So we have a general formula : \[s_\infty = \frac{a}{1-r} \text{ for } |r| < 1\]

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