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.
\[\sum_{i=1}^{\infty} 12\frac{ 3}{ 5}^i-1\]
It's suppose to be 12(3/5)^i-1
This 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.
I got A?
\[ \sum_{i=1}^{\infty} 12\left(\frac{ 3}{ 5}\right)^{i-1} \]?
yes:)
I would change the i=1 to i=0 (and i-1 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
no. It's (3/5)^i-1 not i=1
***no. It's (3/5)^i-1 not i=1*** yes, but I did this: let i= j+1 then i-1 becomes (j+1-1) = 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 \[ \frac{1-r^\infty}{1-r} \]
Ohhh!!! Ok, I get it now! I realise that's what you were doing
I did that so it looks like the "rule" \[ \sum_{i=0}^N r^i = \frac{1-r^N}{1-r} \]
Yes, I agree
r 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 \[ \frac{1-0}{1-\frac{3}{5}} = \frac{1}{\frac{2}{5}}= \frac{5}{2}\] multiply by the leading 12, to get 12*5/2 = 30
divergent 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.
Ok, so we can cross offf C and D?
Then the answer would be A? Right?
choice A says there is no answer. But we can find the answer. (see up above)
Sorry, I meant B :) Since we got 30 as an answer
Thanks so much<3
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