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lala2

  • one year ago

help?!?

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  1. Luigi0210
    • one year ago
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    @Jhannybean

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

  3. Jhannybean
    • one year ago
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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 \[\large \sum_{n-1}^{\infty}-4(-\frac{1}{2})^{n-1} = C\]

  4. lala2
    • one year ago
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    thank you @Luigi0210

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

  6. Luigi0210
    • one year ago
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    that stuff looks so familiar >.<

  7. carlton123
    • one year ago
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    TRue that

  8. lala2
    • one year ago
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    so that is the answer?

  9. carlton123
    • one year ago
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    yes

  10. lala2
    • one year ago
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    oh ok!

  11. lala2
    • one year ago
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    thanks

  12. lala2
    • one year ago
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    you think you could help with another?

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