Results 1 to 4 of 4

Thread: Help with a sequence

  1. #1
    Newbie
    Joined
    Sep 2009
    Posts
    7

    Help with a sequence

    I'm trying to create a formula for this sequence and just can't figure it out:

    { 1/1, 1/3, 1/2, 1/4, 1/3, 1/5, 1/4, 1/6, ... }

    Any ideas?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member redsoxfan325's Avatar
    Joined
    Feb 2009
    From
    Swampscott, MA
    Posts
    943
    Quote Originally Posted by mxrider530 View Post
    I'm trying to create a formula for this sequence and just can't figure it out:

    { 1/1, 1/3, 1/2, 1/4, 1/3, 1/5, 1/4, 1/6, ... }

    Any ideas?
    $\displaystyle a_n=\left\{\begin{array}{lr}\frac{2}{n+1}&:n~odd\\ \frac{2}{n+4}&:n~even\end{array}\right\}$, $\displaystyle n\in\{1,2,3,...\}$
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    12,028
    Thanks
    848
    Hello, mxrider530!

    This is a wicked problem!


    Find a formula for this sequence: . $\displaystyle \frac{1}{1},\;\frac{1}{3},\;\frac{1}{2},\; \frac{1}{4},\; \frac{1}{3},\; \frac{1}{5},\; \frac{1}{4},\; \frac{1}{6},\;\hdots$
    I examined the denominators, $\displaystyle d.$

    . . $\displaystyle \begin{array}{c|cccccccccc}
    n & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ \hline
    d & 1 & 3 & 2 & 4 & 3 & 5 & 4 & 6 & 5 & 7 \end{array}$



    I considered the denominators for odd $\displaystyle n$ and for even $\displaystyle n.$


    $\displaystyle \begin{array}{c|ccccc}
    \text{odd }n & 1 & 3 & 5 & 7 & 9 \\ \hline
    d & 1 & 2 & 3 & 4 & 5 \end{array}$

    . . If $\displaystyle n$ is odd, add 1 and divide by 2: .$\displaystyle d \:=\:\frac{n+1}{2}$ .[1]


    $\displaystyle \begin{array}{c|ccccc}
    \text{even }n & 2 & 4 & 6 & 8 & 10 \\ \hline
    d & 3 & 4 & 5 & 6 & 7 \end{array}$

    . . If $\displaystyle n$ is even, divide by 2 and add 2: .$\displaystyle d \:=\:\frac{n}{2}+2 \:=\:\frac{n+4}{2}$ .[2]



    How can we alternate between the two formulas: .[1] and [2] ?

    . . Multiply [1] by: .$\displaystyle \frac{1-(\text{-}1)^n}{2}\quad\hdots$ multiply [2] by: .$\displaystyle \frac{1 + (\text{-}1)^n}{2} \quad\hdots$ and add the results.


    We have: .$\displaystyle d \;=\;\frac{1-(\text{-}1)^n}{2}\left(\frac{n+1}{2}\right) + \frac{1+(\text{-}1)^n}{2}\left(\frac{n+4}{2}\right)$

    . . This simplifies to: .$\displaystyle d \;=\;\frac{2n+5 + (\text{-}1)^n\cdot3}{4}$



    $\displaystyle \text{Hence, the }n^{th}\text{ fraction is: }\;\;\frac{1}{\dfrac{2n+5 + (\text{-}1)^n\cdot 3}{4}}$

    . . Therefore: .$\displaystyle \boxed{f(n) \;=\;\frac{4}{2n+5 + (\text{-}1)^n\cdot3}} $

    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Sep 2009
    Posts
    7
    Wow, what a need trick to alternate formulas between even and odd! Props
    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

Search Tags


/mathhelpforum @mathhelpforum