1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Find the sum of a geometric progression

  1. Aug 17, 2011 #1
    1. The problem statement, all variables and given/known data

    (1) [itex]\frac{1}{(1+x^{2})}[/itex]+[itex]\frac{1}{(1+x^{2})^{2}}[/itex]+...+[itex]\frac{1}{(1+x^{2})^{n}}[/itex]

    3. The attempt at a solution

    (2) [itex]\frac{1}{(1+x^{2})}[/itex]*[itex]\frac{(1+x^{2})^{n-1}}{(1+x^{2})^{n-1}}[/itex]+[itex]\frac{1}{(1+x^{2})^{2}}[/itex]*[itex]\frac{(1+x^{2})^{n-2}}{(1+x^{2})^{n-2}}[/itex]+...+[itex]\frac{1}{(1+x^{2})^{n}}[/itex]*[itex]\frac{(1+x^{2})^{n-n}}{(1+x^{2})^{n-n}}[/itex]

    (3) [itex]\frac{(1+x^{2})^{n-1}+(1+x^{2})^{n-2}+...+(1+x^{2})^{n-n}}{(1+x^{2})^{n}}[/itex]

    Then I use the formula

    [itex]{S(n)=\frac{n(n-1)}{2}}[/itex]

    I think i didn't do (2) correctly (reduction to a common denominator). Or did I?
     
    Last edited: Aug 17, 2011
  2. jcsd
  3. Aug 17, 2011 #2

    Mentallic

    User Avatar
    Homework Helper

    What you're doing is unnecessary. You already know that the formula for the sum of a geometric progression is

    [tex]S_n=\frac{a(1-r^n)}{1-r}[/tex]

    So what is r and a in this case?
     
  4. Aug 17, 2011 #3
    Yeah I probably should have done it this way.

    Both a and r are: [itex]\frac{1}{1+x^{2}}[/itex]

    Now 3 questions:

    (1) Have I reduced sum to a common denominator correctly (previous post eq. (2))?

    (2) Should I simplify the eq.:

    [tex]S_n=\frac{\frac{1}{1+x^{2}}(1-\frac{1}{1+x^{2}}^n)}{1-\frac{1}{1+x^{2}}}[/tex]

    (3) How to know when to leave the exercise, what is the rule of thumb to remember on how much to simplify?
     
    Last edited: Aug 17, 2011
  5. Aug 17, 2011 #4

    eumyang

    User Avatar
    Homework Helper

    By all means you should simplify. I don't consider a complex fraction to be in simpliest form. Can you simplify so that neither the numerator or denominator contain fractions?
     
  6. Aug 17, 2011 #5

    Mentallic

    User Avatar
    Homework Helper

    Yes you have correctly found the common denominator in (2).

    You should definitely cancel out the [tex]\frac{1}{1+x^2}[/tex] in both the numerator and denominator. From there, I think it would be fine, if you go further as to get rid of the fraction in the numerator then you're going to have two [tex](1+x^2)^n[/tex] terms and some would argue that it isn't as neat.
     
  7. Aug 17, 2011 #6
    [tex]S_n=\frac{\frac{1}{1+x^{2}}(1-\frac{1}{1+x^{2}}^n)}{1-\frac{1}{1+x^{2}}}[/tex]

    I went:

    (1) [tex]S_n=\frac{\frac{1}{1+x^{2}}(1-\frac{1}{1+x^{2}}^n)}{\frac{1+x^{2}}{1+x^{2}}-\frac{1}{1+x^{2}}}[/tex]

    At the end got:

    (2) [tex]S_n=\frac{1-\frac{1}{1+x^{2}}^{n}}{x^{2}}[/tex]

    Can I do smth. more (I doesn't seem to me)? Or should i do something different? Is there a way to get rid of the fraction in the numerator?
     
    Last edited: Aug 17, 2011
  8. Aug 17, 2011 #7

    Mentallic

    User Avatar
    Homework Helper

    In exactly the same way that you would get rid of the fraction in an expression such as this,

    [tex]\frac{1+\frac{1}{x}}{y}[/tex]
     
  9. Aug 17, 2011 #8
    Continuing:

    [tex]S_n=\frac{1-\frac{1}{(1+x^{2})^{n}}}{x^{2}}[/tex]

    Got:

    [tex]S_n=\frac{(1+x^{2})^{n}}{x^{2}(1+x^{2})^{n}-1}[/tex]

    Is this correct?
     
    Last edited: Aug 17, 2011
  10. Aug 18, 2011 #9
    Probably should have written a few more steps:

    (1) [tex]S_n=\frac{1-\frac{1}{(1+x^{2})^{n}}}{x^{2}}[/tex]

    (2) [tex]S_n=\frac{\frac{(1+x^{2})^{n}}{(1+x^{2})^{n}}-\frac{1}{(1+x^{2})^{n}}}{x^{2}}[/tex]

    (3) [tex]S_n=\frac{\frac{(1+x^{2})^{n}-1}{(1+x^{2})^{n}}}{x^{2}}[/tex]

    (4) [tex]S_n=\frac{x^{2}(1+x^{2})^{n}-1}{(1+x^{2})^{n}}[/tex]

    Is this the form that I should leave it in?
     
    Last edited: Aug 18, 2011
  11. Aug 18, 2011 #10
    It should be
    [tex]S_n=\frac{(1+x^{2})^{n}-1}{x^{2}(1+x^{2})^{n}}[/tex]
     
  12. Aug 18, 2011 #11
    Yes, my bad.

    Thank you all for the help.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Find the sum of a geometric progression
  1. Geometric progression (Replies: 3)

  2. Geometric progression (Replies: 2)

  3. Geometric Progression (Replies: 2)

  4. Geometric progression (Replies: 3)

Loading...