# Find the sum of the series

1. Jan 30, 2007

### rcmango

1. The problem statement, all variables and given/known data

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

2. Relevant equations

can use geometric series i think.

3. 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)

Last edited: Jan 30, 2007
2. Jan 30, 2007

### Fredrik

Staff Emeritus
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.

3. Jan 30, 2007

### Fredrik

Staff Emeritus
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.

4. Jan 30, 2007

### rcmango

"Is the first term supposed to be 16/1 " ya 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!

5. Jan 31, 2007

### HallsofIvy

Staff Emeritus
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 7n-1= (7n)(7-1).
$$\frac{4^{n+1}}{7^{n-1}}= \frac{(4)(4^n)}{7^{-1}7^n}$$
$$= \frac{4}{7^{-1}}\frac{4^n}{7^n}= (4)(7)\left(\frac{4}{7}\right)^n= 28\left(\frac{4}{7}\right)^n$$

6. Jan 31, 2007

### 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.

7. Jan 31, 2007

### HallsofIvy

Staff Emeritus
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.

8. Jan 31, 2007

### Schrodinger's Dog

Do you mean this?

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

Last edited: Jan 31, 2007