Results 1 to 3 of 3

Math Help - Simple problem regarding sequences

  1. #1
    Newbie
    Joined
    Feb 2010
    Posts
    5

    Simple problem regarding sequences

    Hello!

    I have a "simple" sequence, but I dont know much about sequences. Is there any easy way of computing what is the nth value of the sequence:

    Sum[e^(-k i)]

    with i going from 0 to n. k is a constant.

    Is there any way of computing this besides doing the actual sum?
    Thanks for all the help
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    Quote Originally Posted by Sosi View Post
    Hello!

    I have a "simple" sequence, but I dont know much about sequences. Is there any easy way of computing what is the nth value of the sequence:

    Sum[e^(-k i)]

    with i going from 0 to n. k is a constant.

    Is there any way of computing this besides doing the actual sum?
    Thanks for all the help
    Hi Sosi,

    you can create a formula as follows

    S=e^0+e^{-k}+e^{-2k}+e^{-3k}+....+e^{-(n-1)k}+e^{-nk}

    =1+e^{-k}\left(e^{-k}+e^{-2k}+....e^{-(n-2)k}+e^{-(n-1)k}\right)<br />
    In brackets is the sum itself with the last term removed

    S=1+e^{-k}\left(S-e^{-nk}\right)

    S=1+Se^{-k}-e^{-(n+1)k}

    S\left(1-e^{-k}\right)=1-e^{-(n+1)k}

    S=\frac{1-e^{-(n+1)k}}{1-e^{-k}}

    Alternatively, you could use the formula for summing a geometric series.

    S=\frac{a\left(1-r^n\right)}{1-r}

    where "a" is the first term, "r" is the common ratio and "n" is the number of terms.

    a=e^0=1

    r=\frac{e^{-2k}}{e^{-k}}=e^{-k}

    There are (n+1) terms as i ranges from 0 to n.

    S=\frac{1\left(1-e^{-(n+1)k}\right)}{1-e^{-k}}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Feb 2010
    Posts
    5
    Quote Originally Posted by Archie Meade View Post
    Hi Sosi,

    you can create a formula as follows

    S=e^0+e^{-k}+e^{-2k}+e^{-3k}+....+e^{-(n-1)k}+e^{-nk}

    =1+e^{-k}\left(e^{-k}+e^{-2k}+....e^{-(n-2)k}+e^{-(n-1)k}\right)<br />
    In brackets is the sum itself with the last term removed

    S=1+e^{-k}\left(S-e^{-nk}\right)

    S=1+Se^{-k}-e^{-(n+1)k}

    S\left(1-e^{-k}\right)=1-e^{-(n+1)k}

    S=\frac{1-e^{-(n+1)k}}{1-e^{-k}}

    Alternatively, you could use the formula for summing a geometric series.

    S=\frac{a\left(1-r^n\right)}{1-r}

    where "a" is the first term, "r" is the common ratio and "n" is the number of terms.

    a=e^0=1

    r=\frac{e^{-2k}}{e^{-k}}=e^{-k}

    There are (n+1) terms as i ranges from 0 to n.

    S=\frac{1\left(1-e^{-(n+1)k}\right)}{1-e^{-k}}
    Wow! Thanks! So S=\frac{1\left(1-e^{-(n+1)k}\right)}{1-e^{-k}} will give me the value that I want. For instance, S(10)=\frac{1\left(1-e^{-(10+1)k}\right)}{1-e^{-k}}

    correct?

    Thanks so much!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 8
    Last Post: September 24th 2011, 08:33 AM
  2. Sequences problem.
    Posted in the Calculus Forum
    Replies: 2
    Last Post: September 19th 2011, 11:28 AM
  3. Simple question regarding sequences and subsequences
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: August 28th 2010, 02:03 AM
  4. Simple sequences problem
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: August 6th 2010, 10:19 AM
  5. Very simple Sequences question
    Posted in the Calculus Forum
    Replies: 2
    Last Post: April 2nd 2010, 07:37 PM

Search Tags


/mathhelpforum @mathhelpforum