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anonymous
 one year ago
Identify whether the series summation of 12 open parentheses 3 over 5 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.
anonymous
 one year ago
Identify whether the series summation of 12 open parentheses 3 over 5 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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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\sum_{i=1}^{\infty} 12\frac{ 3}{ 5}^i1\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0It's suppose to be 12(3/5)^i1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0This is a convergent geometric series. The sum cannot be found. This is a convergent geometric series. The sum is 30. This is a divergent geometric series. The sum cannot be found. This is a divergent geometric series. The sum is 30.

phi
 one year ago
Best ResponseYou've already chosen the best response.2\[ \sum_{i=1}^{\infty} 12\left(\frac{ 3}{ 5}\right)^{i1} \]?

phi
 one year ago
Best ResponseYou've already chosen the best response.2I would change the i=1 to i=0 (and i1 in the exponent to i) just so it looks "more standard" \[ \sum_{i=0}^{\infty} 12\left(\frac{ 3}{ 5}\right)^{i} \\ 12\sum_{i=0}^{\infty} \left(\frac{ 3}{ 5}\right)^{i} \] there is a standard formula for this

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0no. It's (3/5)^i1 not i=1

phi
 one year ago
Best ResponseYou've already chosen the best response.2***no. It's (3/5)^i1 not i=1*** yes, but I did this: let i= j+1 then i1 becomes (j+11) = j and in the summation, the lower limit i=1 becomes j+1= 1 or j=0 and the upper limit stays infinity then I renamed j to i the summation part is \[ \frac{1r^\infty}{1r} \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ohhh!!! Ok, I get it now! I realise that's what you were doing

phi
 one year ago
Best ResponseYou've already chosen the best response.2I did that so it looks like the "rule" \[ \sum_{i=0}^N r^i = \frac{1r^N}{1r} \]

phi
 one year ago
Best ResponseYou've already chosen the best response.2r is 3/5 and we know r*r*r*... gets smaller and smaller, and if we do it enough times, it approaches zero. thus the summation is \[ \frac{10}{1\frac{3}{5}} = \frac{1}{\frac{2}{5}}= \frac{5}{2}\] multiply by the leading 12, to get 12*5/2 = 30

phi
 one year ago
Best ResponseYou've already chosen the best response.2divergent means the sum "diverges" (fancy word for "moves away" ... from a fixed number) in other words, you don't get an answer if it diverges. cross off any answer with "diverges" because we found an answer.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ok, so we can cross offf C and D?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Then the answer would be A? Right?

phi
 one year ago
Best ResponseYou've already chosen the best response.2choice A says there is no answer. But we can find the answer. (see up above)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Sorry, I meant B :) Since we got 30 as an answer
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