Find the sum of the series

[rcmango]

Homework Statement

equation: http://img292.imageshack.us/img292/6239/untitledkl6.jpg

Homework Equations

can use geometric series i think.

The Attempt at a Solution

I think this is a geometric series, i plugged numbers into the equation, starting from 1 and got: 1/16 + 64/7 + 256/49 +...

maybe i can use a formula something like this? --> 1/(1-a)

need help with this please.

 

[Fredrik]

Is the first term supposed to be 16/1 rather than 1/16? Then I would take 16 outside the series. 16*(1+4/7+16/49+...). The stuff in parentheses is the beginning of a geometric series.

 

[Fredrik]

OK, now I've looked at the image too. I suggest you define k=n-1 and rewrite the sum with k as the summation variable instead of n. You will see that you need to move 16 outside the sum to get a geometric series.

 

[rcmango]

"Is the first term supposed to be 16/1 " you it is.

"I suggest you define k=n-1 and rewrite the sum with k as the summation variable instead of n."

..sorry, I'm not too good with the terminology.

however I've attached a new pic, of some of the work i did, but i need help.

http://img262.imageshack.us/img262/4371/untitledme4.jpg

not sure if I'm on the right track, feel free to correct me!

 

[HallsofIvy]

What does that "5" (or is it an "S") in your image mean? The crucial point is that the fractions in your sum are not powers of 4/7. Yes, that is a geometric series- use the formula for the sum of a geometric series.

What Fredrick was talking about before is this- your first image gives
$$\frac{4^{n+1}}{7^{n-1}}$$
and I'm sure you are used to having just "r^n". Okay, take out everything except the "n". 4ⁿ⁺¹= (4ⁿ)(4) and 7^n-1= (7ⁿ)(7^-1).
$$\frac{4^{n+1}}{7^{n-1}}= \frac{(4)(4^n)}{7^{-1}7^n}$$
$$= \frac{4}{7^{-1}}\frac{4^n}{7^n}= (4)(7)\left(\frac{4}{7}\right)^n= 28\left(\frac{4}{7}\right)^n$$

 

[rcmango]

yes, the 5 was an S.

okay, 28(4/7)^n, is that my new S?

where do i go from here? before I would just have a solid fraction and divide through to get what S equals.

 

[HallsofIvy]

yes, the 5 was an S.

okay, 28(4/7)^n, is that my new S?

where do i go from here? before I would just have a solid fraction and divide through to get what S equals.

I have no idea. What do YOU mean by "S"? Perhaps it would be a good idea for you to post what you are using as the formula for the sum of a geometric series.

 

[Schrodinger's Dog]

What does that "5" (or is it an "S") in your image mean? The crucial point is that the fractions in your sum are not powers of 4/7. Yes, that is a geometric series- use the formula for the sum of a geometric series.

What Fredrick was talking about before is this- your first image gives
$$\frac{4^{n+1}}{7^{n-1}}$$
and I'm sure you are used to having just "r^n". Okay, take out everything except the "n". 4ⁿ⁺¹= (4ⁿ)(4) and 7^n-1= (7ⁿ)(7^-1).
$$\frac{4^{n+1}}{7^{n-1}}= \frac{(4)(4^n)}{7^{-1}7^n}$$
$$= \frac{4}{7^{-1}}\frac{4^n}{7^n}= (4)(7)\left(\frac{4}{7}\right)^n= 28\left(\frac{4}{7}\right)^n$$

yes, the 5 was an S.

okay, 28(4/7)^n, is that my new S?

where do i go from here? before I would just have a solid fraction and divide through to get what S equals.

I have no idea. What do YOU mean by "S"? Perhaps it would be a good idea for you to post what you are using as the formula for the sum of a geometric series.

Do you mean this?

$$\sum_{n=1}^\infty \frac{4^{n+1}}{7^{n-1}}= \frac{(4)(4^n)}{7^{-1}7^n}$$ ? Sigma?