# Find the sum of the series

rcmango

## Homework Statement

equation: http://img292.imageshack.us/img292/6239/untitledkl6.jpg [Broken]

## 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)

Last edited by a moderator:

Staff Emeritus
Gold Member
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.

Staff Emeritus
Gold Member
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 [Broken]

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

Last edited by a moderator:
Homework Helper
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).
$$\frac{4^{n+1}}{7^{n-1}}= \frac{(4)(4^n)}{7^{-1}7^n}$$
$$= \frac{4}{7^{-1}}\frac{4^n}{7^n}= (4)(7)\left(\frac{4}{7}\right)^n= 28\left(\frac{4}{7}\right)^n$$

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.

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

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?

Last edited: