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anonymous

  • one year ago

Identify whether the series summation of 15 open parentheses 4 close parentheses to the i minus 1 power from 1 to infinity is a convergent or divergent geometric series and find the sum, if possible.

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  1. anonymous
    • one year ago
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    \[\sum_{\infty}^{i=1}15(4)^i-1\]

  2. anonymous
    • one year ago
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    It's suppose to be 15(4)^i=1

  3. anonymous
    • one year ago
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    This is a convergent geometric series. The sum is –5. This is a divergent geometric series. The sum is –5. This is a convergent geometric series. The sum cannot be found. This is a divergent geometric series. The sum cannot be found.

  4. anonymous
    • one year ago
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    I got A :)

  5. jim_thompson5910
    • one year ago
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    the exponent is `i-1` ??

  6. anonymous
    • one year ago
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    Whether you meant \(5(4)^{i+1}\) or \(5(4)^{i-1}\), consider what happens as \(i\to\infty\). \[\lim_{i\to\infty}5(4)^{i}=\infty\neq0\] What do you know about series of the form \(\sum a_n\) for which \(\lim\limits_{n\to\infty}a_n\neq0\)?

  7. anonymous
    • one year ago
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    Yes the exponent is i-1, and I think it's divergent

  8. anonymous
    • one year ago
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    That's right.

  9. anonymous
    • one year ago
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    And I couldnt find the sum so D!

  10. jim_thompson5910
    • one year ago
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    Yep whenever it's divergent, the infinite number of terms don't sum to one fixed number.

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