Find the sum of the series

  • Thread starter rcmango
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  • #1
rcmango
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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)

need help with this please.
 
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Answers and Replies

  • #2
Fredrik
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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
Fredrik
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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
rcmango
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"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!
 
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  • #5
HallsofIvy
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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
[tex]\frac{4^{n+1}}{7^{n-1}}[/tex]
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).
[tex]\frac{4^{n+1}}{7^{n-1}}= \frac{(4)(4^n)}{7^{-1}7^n}[/tex]
[tex]= \frac{4}{7^{-1}}\frac{4^n}{7^n}= (4)(7)\left(\frac{4}{7}\right)^n= 28\left(\frac{4}{7}\right)^n[/tex]
 
  • #6
rcmango
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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
HallsofIvy
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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.
 
  • #8
Schrodinger's Dog
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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
[tex]\frac{4^{n+1}}{7^{n-1}}[/tex]
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).
[tex]\frac{4^{n+1}}{7^{n-1}}= \frac{(4)(4^n)}{7^{-1}7^n}[/tex]
[tex]= \frac{4}{7^{-1}}\frac{4^n}{7^n}= (4)(7)\left(\frac{4}{7}\right)^n= 28\left(\frac{4}{7}\right)^n[/tex]

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?

[tex]\sum_{n=1}^\infty \frac{4^{n+1}}{7^{n-1}}= \frac{(4)(4^n)}{7^{-1}7^n}[/tex] ? Sigma?
 
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