Results 1 to 3 of 3

Math Help - help with fraction sequences

  1. #1
    Newbie
    Joined
    Oct 2008
    From
    London UK
    Posts
    1

    Exclamation help with fraction sequences

    i need help with these problems coz my maths teacher has my book i dont know wat 2 do so cud u plz explain and show me how 2 do it. thanks muchly

    *1 write an expression for the nth term of the sequence:
    1/3 , 2/5 , 3/7 , 4,9 , 5/11 ..............


    *2 the nth term is n/n^2+1

    the 1st term in the sequence is 1/2
    what are the next 3 terms?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member
    Joined
    Oct 2008
    Posts
    35
    help with fraction sequences please xox

    The Term n :Tn of an arithmetic sequence can be given by
    Tn = a + (n-1)d
    where n is the nth term
    a is the first term and d is the common difference ( ie any term less the previous term)

    so T1 = a + (1-1)d=a
    T2 = a +(2-1)d= a+d
    T3 = a + (3-1)d = a+2d and so on

    In your problem note that the numerator of any Tn is n

    The denominator : first term is 3 ; common difference is 2 so denominator is Tn = a + (n-1)d = 3 +2(n-1)

    numerator/denominator Tn = n/( 3 + 2(n-1))


    *2 Substitution


    Tn = n/ (n^2 +1)
    then for T1 replace each n with 1 : 1/(1^2 +1 ) = 1/2
    similarly for next 3 terms T2, T3 and T4
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,908
    Thanks
    766
    Hello, rachel27!

    Are you really having trouble with these?


    Write an expression for the n^{th} term of the sequence:

    . . \frac{1}{3}\quad \frac{2}{5}\quad \frac{3}{7}\quad\frac{4}{9}\quad\frac{5}{11}\quad\  hdots
    Can you see the pattern of the numerators and denominators?

    . . \begin{array}{c||c|c|c|c|c}<br />
n & 1 & 2 & 3 & 4 & 5 \\ \hline \hline<br />
\text{num'r} & 1 & 2 & 3 & 4 & 5 \\ \hline<br />
\text{den'r} & 3 & 5 & 7 & 9 & 11 \\ \hline \end{array}


    It should be obvious . . .

    . . The n^{th} numerator is n.

    . . The n^{th} denominator is: 2n+1

    Therefore, the n^{th} fraction is: . \frac{n}{2n+1}




    2) The n^{th} term is \frac{n}{n^2+1}

    The first term is \frac{1}{2}

    What are the next 3 terms?
    How hard can this be?


    f(n) \:=\:\frac{n}{n^2+1}

    . . \begin{array}{cccccc}f(1) &=&\dfrac{1}{1^2+1} &=& \dfrac{1}{2} \\ \\[-3mm]<br /> <br />
f(2) &=& \dfrac{2}{2^2+1} &=& \dfrac{2}{5} \\ \\[-3mm]<br /> <br />
f(3) &=& \dfrac{3}{3^2+1} &=& \dfrac{3}{10} \\ \\[-3mm]<br /> <br />
f(4) &=& \dfrac{4}{4^2+1} &=& \dfrac{4}{17} \end{array}

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. partial fraction from improper fraction
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 24th 2010, 03:28 AM
  2. Replies: 2
    Last Post: September 28th 2009, 08:18 PM
  3. Monotone sequences and Cauchy sequences
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: March 21st 2009, 09:59 PM
  4. Replies: 5
    Last Post: January 16th 2008, 05:51 PM
  5. Replies: 4
    Last Post: September 1st 2007, 12:05 AM

Search Tags


/mathhelpforum @mathhelpforum