Results 1 to 7 of 7
Like Tree2Thanks
  • 1 Post By Plato
  • 1 Post By Idea

Thread: sequence

  1. #1
    Junior Member
    Joined
    Jun 2016
    From
    Brasil
    Posts
    69

    sequence

    Let an be the n-th term of the following sequence 1/1, 1/4, 3/4, 1/9, 3/9, 5/9, 1/16, 3/16, 5/16, 7/16, 1/25...


    Find the maximal n satisfying an >= 1/10.

    The answer is 190


    I do not know how to ask that question
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Feb 2015
    From
    Ottawa Ontario
    Posts
    1,654
    Thanks
    310

    Re: sequence

    Quote Originally Posted by provasanteriores View Post
    Let an be the n-th term of the following sequence:
    1/1, 1/4, 3/4, 1/9, 3/9, 5/9, 1/16, 3/16, 5/16, 7/16, 1/25...

    Find the maximal n satisfying an >= 1/10.
    The answer is 190.

    I do not know how to ask that question
    And I do not know WHY you ask that question!
    "an" should be shown as a(n): ok?
    And the denominators are squares: ok?

    Terms : number of terms (number of terms so far)
    1/1^2 : 1 (1)
    1/2^2, 3/2^2 : 2 (3)
    1/3^2, 3/3^2, 5/3^2 : 3 (6)
    and so on...
    So what you have is "sum of natural numbers" so far...

    Denominator 13^2 ends with 19/13^2 which is the 220th term.
    (1 + 3 + 6 + 10 + 15 + 21 + 28 + 36 + 45 + 55 = 220).
    So at 220th term, we have 19/13^2 = .11242....; still > 1/10.

    The next bunch ends with 21/15^2 which is 286th term:
    21/15^2 = .09333....
    So your answer lies somewhere in that group: GO FIND IT!!

    I'm not doing any more for you...got a headache!!
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Aug 2006
    Posts
    21,274
    Thanks
    2644
    Awards
    1

    Re: sequence

    Quote Originally Posted by provasanteriores View Post
    Let an be the n-th term of the following sequence
    1/1, 1/4, 3/4, 1/9, 3/9, 5/9, 1/16, 3/16, 5/16, 7/16, 1/25...
    Find the maximal n satisfying an >= 1/10.
    The answer is 190
    Define ${a_n} = \left\lfloor {\sqrt {2n} + 0.5} \right\rfloor$ then $(a_n)^2$ is the denominator of the term of the sequence $T_n$.
    However, the numerators are a bit trickery to find.

    $T_{16}=\dfrac{1}{36}~T_{17}=\dfrac{3}{36}~\cdots~ T_{21}=\dfrac{11}{36}$
    Thanks from DenisB
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Feb 2015
    From
    Ottawa Ontario
    Posts
    1,654
    Thanks
    310

    Re: sequence

    ...and the sum of each "group" = 1
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member
    Joined
    Jun 2013
    From
    Lebanon
    Posts
    709
    Thanks
    327

    Re: sequence

    Write the elements of the sequence in the form of a lower triangular matrix (zeroes above the diagonal)

    Row n = \frac{2k-1}{n^2} where 1\leq k\leq n followed by a bunch of zeroes

    The diagonal entry is the largest number in its row

    so we need \frac{2n-1}{n^2}<\frac{1}{10} which gives n\geq 20

    The first 19 rows have a total of 1+2+...+19=190 entries (not counting the zeroes)

    Check: \frac{37}{19^2}>\frac{1}{10} and \frac{39}{20^2}<\frac{1}{10}
    Last edited by Idea; Jul 16th 2017 at 06:48 AM.
    Thanks from DenisB
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Joined
    Feb 2015
    From
    Ottawa Ontario
    Posts
    1,654
    Thanks
    310

    Re: sequence

    Nice "Idea"!

    I made a silly goof in my post...
    at numerator 15, started jumping by 2 instead of 1...
    I'm standing in the corner!

    Denominator 19^2 ends with 37/19^2 which is the 190th term.
    So at 190th term, we have 37/19^2 = .1025....; still > 1/10.

    The next bunch ends with 39/20^2 which is 210th term:
    39/20^2 = .0975

    So 190 is the man!
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor
    Joined
    Feb 2015
    From
    Ottawa Ontario
    Posts
    1,654
    Thanks
    310

    Re: sequence

    Quote Originally Posted by Idea View Post
    The diagonal entry is the largest number in its row
    Or the largest number in a row is the last one;
    like 9 / 5^2 for row#5.

    (2u + 1) / [(2u +2)/2]^2 = 1 / 10
    u = ~18.49
    u = CEIL(u) = 19
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: Aug 24th 2010, 02:10 AM
  2. Replies: 0
    Last Post: Jul 4th 2010, 12:05 PM
  3. Replies: 2
    Last Post: Mar 1st 2010, 11:57 AM
  4. sequence membership and sequence builder operators
    Posted in the Discrete Math Forum
    Replies: 0
    Last Post: Jun 4th 2009, 03:16 AM
  5. Replies: 12
    Last Post: Nov 15th 2006, 12:51 PM

/mathhelpforum @mathhelpforum