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!

Please help with simple geometric series

  1. Apr 26, 2006 #1
    how does this come out to this?


    i dont understand

  2. jcsd
  3. Apr 26, 2006 #2


    User Avatar
    Homework Helper
    Gold Member

    What part do you not understand?
  4. Apr 26, 2006 #3


    User Avatar
    Homework Helper

    What part don't you get ? It's just algebraic rearrangement :

    [tex]2^{2n+1}5^{-n} = 2^{2n}(2)(5^{-n}) = (2)({(2^2)}^n)(5^{-n}) = (2)(4^n)(\frac{1}{5^n}) = (2)(\frac{4^n}{5^n}) = (2){(\frac{4}{5})}^n[/tex]

    I would've just left it there since it's obvious this is a geometric series, but the text apparently felt it necessary to take another step (which seems rather pointless), but it's also just algebra.

    [tex](2){(\frac{4}{5})}^n = (2)(\frac{4}{5}){(\frac{4}{5})}^{n-1} = (\frac{8}{5}){(\frac{4}{5})}^{n-1}[/tex]

    Clear enough ?
  5. Apr 26, 2006 #4


    User Avatar
    Staff Emeritus
    Science Advisor

    22n+1=(2)22n= (2)(22)n= (2)(4n).
    5-n= 1/5n

    22n+15-n= (2)(4n)/5n= (2)(4/5)n.

    Curious3141, the reason for "next step" is that the sum starts at n= 1. The standard formula for the sum of a geometric series,
    [tex]\Sigma_{n=0}^\infty ar^n= \frac{a}{1-r}[/tex]
    requires that the sum start at n=0.
    [tex]\Sigma_{n=1}^\infty\(\frac{8}{5}\)\(\frac{4}{5}\)^{n-1}= \Sigma_{n=0}^\infty\(\frac{8}{5}\)\({4}{5}\)^n[/tex]
    Last edited: Apr 26, 2006
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Please help with simple geometric series