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Some sums, don't sum up

  1. Jul 20, 2006 #1
    Some sums, don't sum up :)

    I have a problem that require some math tricks, and after I tried to solve it myself I looked at the answer and I don't understand how this is done :
    [tex]
    \[
    \sum\limits_{k = 0}^n {\left( {\frac{2}{5}} \right)^k } + \sum\limits_{k = 0}^n {\left( {\frac{3}{5}} \right)^k } = \frac{5}{3}\left( {1 - \left( {\frac{2}{5}} \right)^{n + 1} } \right) + \frac{5}{2}\left( {1 - \left( {\frac{3}{5}} \right)^{n + 1} } \right)
    \]
    [/tex]

    An advice pls, thx!
     
  2. jcsd
  3. Jul 20, 2006 #2

    nazzard

    User Avatar
    Gold Member

    Hello seaglespn,

    Hint: geometric series :wink:

    [tex]\sum\limits_{k = 0}^n q^{k}=\frac{1-q^{n+1}}{1-q}[/tex]

    for [tex]|q|<1[/tex]

    Do you know how to prove this identity?



    Although not necessary for solving this problem you might want to take a look at the infinite geometric series as well.

    [tex]\sum\limits_{k = 0}^\infty q^{k}[/tex]

    for [tex]|q|<1[/tex]

    What would be the limit?

    Regards,

    nazzard
     
    Last edited: Jul 20, 2006
  4. Jul 20, 2006 #3

    Ok, I have done the math, and I end up with the correct answer, after I wasn't so sure about the : [tex] \[
    b_n = b_1 \frac{{q^n - 1}}{{q - 1}}
    \]
    [/tex]

    Where the power of q must be the TOTAL number of elements...

    Code (Text):

    Sorry, my mistake... :smile:
    The sum thends to a constant.... but that might be a definitions somewhere....
    And it didn't rings any bell to me...
    A constant "variable" due to q. :smile: .
    Goofy me...
     
    Thx for the help!

    Regards,
    seaglespn.
     
    Last edited: Jul 20, 2006
  5. Jul 20, 2006 #4

    nazzard

    User Avatar
    Gold Member

    Try again please :smile:

    [tex]\sum\limits_{k = 0}^\infty q^{k}=\lim_{\substack{n\rightarrow\infty}}\sum\limits_{k = 0}^{n} q^{k}=\lim_{\substack{n\rightarrow\infty}}\frac{1-q^{n+1}}{1-q}=?[/tex]

    Remember [tex]|q|<1[/tex]
     
    Last edited: Jul 20, 2006
  6. Jul 20, 2006 #5
    [tex]
    \[
    \mathop {\lim }\limits_{n \to \infty } \frac{{1 - q^{n + 1} }}{{1 - q}} = \frac{1}{{1 - q}} = ?
    \]
    [/tex]
     
  7. Jul 20, 2006 #6
    [tex]
    \[
    \mathop {\lim }\limits_{n \to \infty } \frac{{1 - q^{n + 1} }}{{1 - q}} = \frac{1}{{1 - q}} \]

    [/tex]
    ?
    Sorry about double post... my refresh is kinda slow :smile:
     
  8. Jul 20, 2006 #7

    nazzard

    User Avatar
    Gold Member

    correct :smile:

    Regards,

    nazzard
     
  9. Jul 20, 2006 #8
    Thanks for your help @nazzard... :smile:

    Cheers!
     
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