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anonymous
 one year ago
Find the sum of the infinite series.
The sum(sigma) from k = 1 to infinity of 11 times two thirds to the k power.
A. 44/3
B. 22
C. 33/5
D. 22/5
anonymous
 one year ago
Find the sum of the infinite series. The sum(sigma) from k = 1 to infinity of 11 times two thirds to the k power. A. 44/3 B. 22 C. 33/5 D. 22/5

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freckles
 one year ago
Best ResponseYou've already chosen the best response.1sounds like you have a geometric series.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the answer i got was 33, but that's not one of my answers so im a little confused

freckles
 one year ago
Best ResponseYou've already chosen the best response.1so I will probably go with the formula to find the sum of the geometric series... which is: \[\sum_{n=1}^{\infty} a r^{n1}=a \frac{1}{1r} \text{ where } r<1 \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeah i did that one and i got 33 so idk

freckles
 one year ago
Best ResponseYou've already chosen the best response.1you shouldn't get 33 we can manipulate your sum to fit our form: you have \[\sum_{k=1}^{\infty}11 (\frac{2}{3})^k \\ \text{ \let } k=n1 \\ \text{ so we have} \\ \sum_{n1=1}^{\infty}11(\frac{2}{3})^{n1}=\sum_{n=2}^{\infty}11(\frac{2}{3})^{n1} \\ =\sum_{n=1}^{\infty}11(\frac{2}{3})^{n1}11(\frac{2}{3})^{11}\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.1there is another way you could have manipulated the form to get the form we wanted: \[\sum_{k=1}^{\infty}11(\frac{2}{3})^{k} \\ =\sum_{k=1}^{\infty}11(\frac{2}{3})(\frac{2}{3})^{1}(\frac{2}{3})^{k} \\ =\sum_{k=1}^{\infty}11(\frac{2}{3})(\frac{2}{3})^{k1} \\ \sum_{k=1}^{\infty}\frac{22}{3}(\frac{2}{3})^{k1}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i understand what you did in the first part, but it started from n=2. why did you start from n=1 and subtract the second part?

myininaya
 one year ago
Best ResponseYou've already chosen the best response.1He started in a zero

myininaya
 one year ago
Best ResponseYou've already chosen the best response.1\[\sum_{n=2}^{\infty}11(\frac{2}{3})^{k1} \\ 11(\frac{2}{3})^{11}+11(\frac{2}{3})^{11}+\sum_{n=2}^{\infty}11(\frac{2}{3})^{k1} \\ 11(\frac{2}{3})^{11}+\sum_{n=1}^{\infty}11(\frac{2}{3})^{k1}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i got 22 for my answer when i redid it and it was right. ohh ok that makes sense now.

myininaya
 one year ago
Best ResponseYou've already chosen the best response.1I kind of like the second way he did it

myininaya
 one year ago
Best ResponseYou've already chosen the best response.1he multiply a 1 in the 1 being of the form: \[(\frac{2}{3})(\frac{2}{3})^{1}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeah i think its a good way of looking at it. thank you guys
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