Results 1 to 3 of 3

Math Help - Determining the general term

  1. #1
    Junior Member
    Joined
    Feb 2011
    Posts
    56

    Determining the general term

    The general term is given by 3r+1+(-1)^r+1

    So by substituting r=0,1,2,3.... I get a sequence like this: 0, 5, 6, 11, 12, 17, 18.....

    It seems to form some pattern. So i wonder, can i deduce the general term with only the sequence? How?

    Please kindly elaborate more if possible because i really keen to learn how.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor alexmahone's Avatar
    Joined
    Oct 2008
    Posts
    1,074
    Thanks
    7

    Re: Determining the general term

    Quote Originally Posted by MichaelLight View Post
    The general term is given by 3r+1+(-1)^r+1

    So by substituting r=0,1,2,3.... I get a sequence like this: 0, 5, 6, 11, 12, 17, 18.....

    It seems to form some pattern. So i wonder, can i deduce the general term with only the sequence? How?

    Please kindly elaborate more if possible because i really keen to learn how.
    The sequence can be described as follows:
    Consider the rth term (r starts from 0).
    If r is even, a_r = 3r.
    If r is odd, a_r = 3r+2.
    So, we need to subtract 1 from 3r+1 if r is even and add 1 to 3r+1 if r is odd. An easy way to do this would be to add (-1)^{r+1} to 3r+1.
    Thus, we find that the general term is given by a_r=3r+1+(-1)^{r+1}.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,738
    Thanks
    645

    Re: Determining the general term

    Hello, MichaelLight!

    The general term is given by: . a_n \:=\: 3n+1+(\text{-}1)^{n+1}

    So by substituting r\:=\:0,1,2,3\,\hdots I get a sequence like this: 0, 5, 6, 11, 12, 17, 18 \hdots

    It seems to form some pattern.
    So i wonder, can i deduce the general term from the sequence?

    When given an increasing sequence of terms, I usually take the difference
    . . of consecutive terms. Then the differences of the differences, and so on.
    We hope to find that the n^{th} differences are constant.
    . . This indicates that the generating function is an n^{th} degree polynomial.


    So we have:

    . . \begin{array}{c|cccccccccccccc} \text{Sequence} & 0 && 5 && 6 && 1 && 12 && 17 && 18 \\  \text{1st diff.} && 5 && 1 && 5 && 1 && 5 && 1 \\ \text{2nd diff.} &&& \text{-}4 && 4 && \text{-}4 && 4 && \text{-}4 \end{array}

    But we see that the 1st difference has alternating terms.
    This indicates that the sequence is composed of two subsequences,
    . . one for "even" terms, one for "odd" terms.


    We have:

    . . \begin{array}{c||c|c|c|c|c|c|c|c|}n & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline \text{even} & 0 && 6 && 12 && 18 \\ \hline \text{odd} && 5 && 11 && 17 & \\ \hline \end{array}


    We see that, for even n:\;a_n \:=\:3n
    . . . . . . . . . . for odd n:\;a_n \:=\:3n+2

    And from here, we can consult Alex's excellent explanation.

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. how to determine the general term
    Posted in the Algebra Forum
    Replies: 5
    Last Post: April 8th 2010, 11:11 AM
  2. General term
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 10th 2010, 11:21 PM
  3. general term
    Posted in the Algebra Forum
    Replies: 2
    Last Post: September 15th 2009, 03:00 AM
  4. general term
    Posted in the Algebra Forum
    Replies: 5
    Last Post: September 2nd 2009, 07:11 AM
  5. Replies: 7
    Last Post: August 31st 2007, 08:18 PM

Search Tags


/mathhelpforum @mathhelpforum