## 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

1. freckles

sounds like you have a geometric series.

2. anonymous

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

3. freckles

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

4. anonymous

yeah i did that one and i got 33 so idk

5. freckles

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

6. freckles

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

7. anonymous

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?

8. myininaya

He started in a zero

9. myininaya

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

10. myininaya

11. anonymous

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

12. myininaya

I kind of like the second way he did it

13. myininaya

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

14. anonymous

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