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lala2

help?!?

  • 10 months ago
  • 10 months ago

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  1. Luigi0210
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    @Jhannybean

    • 10 months ago
  2. Luigi0210
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    I have never gotten to series in calculus Lala, sorry.. but I called someone better for you :)

    • 10 months ago
  3. Jhannybean
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    In a geometric series, the series will converge if \[\large |r|<1\]In this case, \[|-\frac{1}{2}| <1 \implies \frac{1}{2}< 1\] therefore the \[\large \sum_{n-1}^{\infty}-4(-\frac{1}{2})^{n-1} = C\]

    • 10 months ago
  4. lala2
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    thank you @Luigi0210

    • 10 months ago
  5. Jhannybean
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    Because it converges, your sum : \[\large S=\frac{a}{1-r}\] where \[\large r=-\frac{1}{2}\]

    • 10 months ago
  6. Luigi0210
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    that stuff looks so familiar >.<

    • 10 months ago
  7. carlton123
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    TRue that

    • 10 months ago
  8. lala2
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    so that is the answer?

    • 10 months ago
  9. carlton123
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    yes

    • 10 months ago
  10. lala2
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    oh ok!

    • 10 months ago
  11. lala2
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    thanks

    • 10 months ago
  12. lala2
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    you think you could help with another?

    • 10 months ago
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