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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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

Mathematics
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sounds like you have a geometric series.
the answer i got was 33, but that's not one of my answers so im a little confused
so I will probably go with the formula to find the sum of the geometric series... which is: \[\sum_{n=1}^{\infty} a r^{n-1}=a \frac{1}{1-r} \text{ where } |r|<1 \]

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yeah i did that one and i got 33 so idk
you shouldn't get 33 we can manipulate your sum to fit our form: you have \[\sum_{k=1}^{\infty}11 (\frac{2}{3})^k \\ \text{ \let } k=n-1 \\ \text{ so we have} \\ \sum_{n-1=1}^{\infty}11(\frac{2}{3})^{n-1}=\sum_{n=2}^{\infty}11(\frac{2}{3})^{n-1} \\ =\sum_{n=1}^{\infty}11(\frac{2}{3})^{n-1}-11(\frac{2}{3})^{1-1}\]
there is another way you could have manipulated the form to get the form we wanted: \[\sum_{k=1}^{\infty}11(\frac{2}{3})^{k} \\ =\sum_{k=1}^{\infty}11(\frac{2}{3})(\frac{2}{3})^{-1}(\frac{2}{3})^{k} \\ =\sum_{k=1}^{\infty}11(\frac{2}{3})(\frac{2}{3})^{k-1} \\ \sum_{k=1}^{\infty}\frac{22}{3}(\frac{2}{3})^{k-1}\]
i 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?
He started in a zero
\[\sum_{n=2}^{\infty}11(\frac{2}{3})^{k-1} \\ -11(\frac{2}{3})^{1-1}+11(\frac{2}{3})^{1-1}+\sum_{n=2}^{\infty}11(\frac{2}{3})^{k-1} \\ -11(\frac{2}{3})^{1-1}+\sum_{n=1}^{\infty}11(\frac{2}{3})^{k-1}\]
he added in a zero*
i got 22 for my answer when i redid it and it was right. ohh ok that makes sense now.
I kind of like the second way he did it
he multiply a 1 in the 1 being of the form: \[(\frac{2}{3})(\frac{2}{3})^{-1}\]
yeah i think its a good way of looking at it. thank you guys

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