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Find the sum of the series

  1. Jan 30, 2007 #1
    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)

    need help with this please.
     
    Last edited: Jan 30, 2007
  2. jcsd
  3. Jan 30, 2007 #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.
     
  4. Jan 30, 2007 #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.
     
  5. Jan 30, 2007 #4
    "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!
     
  6. Jan 31, 2007 #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]
     
  7. Jan 31, 2007 #6
    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.
     
  8. Jan 31, 2007 #7

    HallsofIvy

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    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.
     
  9. Jan 31, 2007 #8
    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?
     
    Last edited: Jan 31, 2007
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