Results 1 to 5 of 5

Thread: Geometric Progression and Summation

  1. #1
    Super Member
    Joined
    Dec 2009
    Posts
    755

    Geometric Progression

    The sum of the first $\displaystyle n$ terms of a series is given by the expression $\displaystyle (1-3^{-2n})$.

    Find the least value of $\displaystyle k$ such that the sum of the terms from the $\displaystyle k^{th}$ term is less than $\displaystyle \frac{1}{3000}$

    $\displaystyle 1-3^{2k} < \frac{1}{3000}$

    $\displaystyle -3^{-2k} < -\frac{2999}{3000}$

    $\displaystyle 3^{-2k} > \frac{2999}{3000}$

    $\displaystyle -2k < \frac{lg\frac{2999}{3000}}{lg3}$

    $\displaystyle k > 0.0001517$

    but ans is 5
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    12,880
    Thanks
    1946
    It doesn't say to sum the terms up to the $\displaystyle \displaystyle k^{\textrm{th}}$ term, it says FROM the $\displaystyle \displaystyle k^{\textrm{th}}$ term, presumably up to the $\displaystyle \displaystyle n^{\textrm{th}}$ term.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Dec 2009
    Posts
    755
    Quote Originally Posted by Prove It View Post
    It doesn't say to sum the terms up to the $\displaystyle \displaystyle k^{\textrm{th}}$ term, it says FROM the $\displaystyle \displaystyle k^{\textrm{th}}$ term, presumably up to the $\displaystyle \displaystyle n^{\textrm{th}}$ term.
    thanks for pointing that out! i'll attempt the question again with this in mind.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    Joined
    Dec 2009
    Posts
    755
    $\displaystyle \sum_{n = k}^N(1-3^{-2n}) < \frac{1}{3000}$

    $\displaystyle \sum_{n = k}^N1-\sum_{n = k}^N(3^{-2n}) < \frac{1}{3000}$

    $\displaystyle (N-k+1)(1)-(3^{-2k}+3^{-2(k+1)}+...+3^{-2N}) < \frac{1}{3000}$

    $\displaystyle N-k+1-\frac{3^{-2k}(1-9^{N-k+1})}{1-9}<\frac{1}{3000}$

    but N doesnt seem to be cancelling out
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member
    Joined
    Dec 2009
    Posts
    755
    Quote Originally Posted by Punch View Post
    $\displaystyle \sum_{n = k}^N(1-3^{-2n}) < \frac{1}{3000}$

    $\displaystyle \sum_{n = k}^N1-\sum_{n = k}^N(3^{-2n}) < \frac{1}{3000}$

    $\displaystyle (N-k+1)(1)-(3^{-2k}+3^{-2(k+1)}+...+3^{-2N}) < \frac{1}{3000}$

    $\displaystyle N-k+1-\frac{3^{-2k}(1-9^{N-k+1})}{1-9}<\frac{1}{3000}$

    but N doesnt seem to be cancelling out
    am i on the right track?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Geometric Progression or Geometric Series
    Posted in the Math Topics Forum
    Replies: 8
    Last Post: Oct 8th 2009, 07:31 AM
  2. Replies: 8
    Last Post: Mar 23rd 2009, 07:26 AM
  3. Geometric progression
    Posted in the Geometry Forum
    Replies: 3
    Last Post: Jun 13th 2008, 06:25 PM
  4. geometric progression
    Posted in the Algebra Forum
    Replies: 2
    Last Post: Apr 18th 2008, 09:55 AM
  5. geometric progression
    Posted in the Algebra Forum
    Replies: 5
    Last Post: Feb 24th 2007, 02:46 PM

Search Tags


/mathhelpforum @mathhelpforum