anonymous
  • anonymous
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.
Mathematics
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
\[\sum_{\infty}^{i=1}15(4)^i-1\]
anonymous
  • anonymous
It's suppose to be 15(4)^i=1
anonymous
  • anonymous
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.

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anonymous
  • anonymous
I got A :)
jim_thompson5910
  • jim_thompson5910
the exponent is `i-1` ??
anonymous
  • anonymous
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 \(\lim\limits_{n\to\infty}a_n\neq0\)?
anonymous
  • anonymous
Yes the exponent is i-1, and I think it's divergent
anonymous
  • anonymous
That's right.
anonymous
  • anonymous
And I couldnt find the sum so D!
jim_thompson5910
  • jim_thompson5910
Yep whenever it's divergent, the infinite number of terms don't sum to one fixed number.

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