Question

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

Answer 1

sounds like you have a geometric series.

Answer 2

the answer i got was 33, but that's not one of my answers so im a little confused

Answer 3

so I will probably go with the formula to find the sum of the geometric series... which is:
\[\sum_{n=1}^{\infty} a r^{n-1}=a \frac{1}{1-r} \text{ where } |r|<1 \]

Answer 4

yeah i did that one and i got 33 so idk

Answer 5

you shouldn't get 33
we can manipulate your sum to fit our form:
you have
\[\sum_{k=1}^{\infty}11 (\frac{2}{3})^k \\ \text{ \let } k=n-1 \\ \text{ so we have} \\ \sum_{n-1=1}^{\infty}11(\frac{2}{3})^{n-1}=\sum_{n=2}^{\infty}11(\frac{2}{3})^{n-1} \\ =\sum_{n=1}^{\infty}11(\frac{2}{3})^{n-1}-11(\frac{2}{3})^{1-1}\]

Answer 6

there is another way you could have manipulated the form to get the form we wanted:
\[\sum_{k=1}^{\infty}11(\frac{2}{3})^{k} \\ =\sum_{k=1}^{\infty}11(\frac{2}{3})(\frac{2}{3})^{-1}(\frac{2}{3})^{k} \\ =\sum_{k=1}^{\infty}11(\frac{2}{3})(\frac{2}{3})^{k-1} \\ \sum_{k=1}^{\infty}\frac{22}{3}(\frac{2}{3})^{k-1}\]

Answer 7

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?

Answer 8

He started in a zero

Answer 9

\[\sum_{n=2}^{\infty}11(\frac{2}{3})^{k-1} \\ -11(\frac{2}{3})^{1-1}+11(\frac{2}{3})^{1-1}+\sum_{n=2}^{\infty}11(\frac{2}{3})^{k-1} \\ -11(\frac{2}{3})^{1-1}+\sum_{n=1}^{\infty}11(\frac{2}{3})^{k-1}\]

Answer 10

he added in a zero*

Answer 11

i got 22 for my answer when i redid it and it was right. ohh ok that makes sense now.

Answer 12

I kind of like the second way he did it

Answer 13

he multiply a 1 in
the 1 being of the form:
\[(\frac{2}{3})(\frac{2}{3})^{-1}\]

Answer 14

yeah i think its a good way of looking at it. thank you guys