Results 1 to 4 of 4

Math Help - Find A And B

  1. #1
    Member
    Joined
    Nov 2006
    From
    chicago
    Posts
    156

    Find A And B

    Find A and B so that 1 / [(4n-3)(4n+1)] = A / (4n-3) + B / (4n+1).





    Thanks for your time, efforts, and help!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Quote Originally Posted by Mr_Green View Post
    Find A and B so that 1 / [(4n-3)(4n+1)] = A / (4n-3) + B / (4n+1).
    You have,
    \frac{1}{(4n-3)(4n+1)}=\frac{A}{4n-3}+\frac{B}{4n+1}
    Multiply through by,
    (4n-3)(4n+1),
    1=A(4n+1)+B(4n-3)
    1=n(4A+4B)+(A-3B)
    We have,
    4A+4B=0
    A-3B=1
    Solve, to get,
    A=1/4,B=-1/4
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Apr 2005
    Posts
    1,631
    Quote Originally Posted by Mr_Green View Post
    Find A and B so that 1 / [(4n-3)(4n+1)] = A / (4n-3) + B / (4n+1).





    Thanks for your time, efforts, and help!
    Here is one way.

    1 /[(4n-3)(4n+1)] = A/(4n-3) +B/(4n+1)
    Multiply both sides by (4n-3)(4n+1)
    1 = A(4n+1) +B(4n-3) ---------------(i)

    So that B is eliminated, make (4n-3) equal to zero.
    4n -3 = 0
    n = 3/4
    So, when n=3/4, in (i),
    1 = A(4(3/4) +1) +B(4(3/4) -3)
    1 = A(3+1) +B(3-3)
    1 = 4A
    A = 1/4 ----------------------answer.

    So that A is eliminated, make (4n+1) equal to zero.
    4n +1 = 0
    n = -1/4
    So, when n = -1/4, in (i),
    1 = A(4(-1/4) +1) +B(4(-1/4) -3)
    1 = A(-1 +1) +B(-1 -3)
    1 = -4B
    B = -1/4 ----------------------answer.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,865
    Thanks
    744
    Hello, Mr_Green!

    Another method . . .


    Find A and B so that: . \frac{1}{(4n-3)(4n+1)} \:= \:\frac{A}{4n-3} + \frac{B}{4n+1}
    Multiply through by (4n-3)(4n+1)\!:\;\;1 \;=\;(4n+1)A + (4n-3)B


    Let n = \frac{3}{4}\!:\;\;1 \:=\:\left[4\left(\frac{3}{4}\right) + 1\right]A + \left[4\left(\frac{3}{4}\right) - 3\right] B

    . . 1 \:=\:4\!\cdot\!A + 0\!\cdot\!B\quad\Rightarrow\quad\boxed{ A\,=\,\frac{1}{4}}


    Let n = -\frac{1}{4}\!:\;\;1\:=\:\left[4\left(-\frac{1}{4}\right) + 1\right]A + \left[4\left(-\frac{1}{4}\right) - 3\right]B

    . . 1 \:=\:0\!\cdot\!A - 4\!\cdot\!B\quad\Rightarrow\quad\boxed{ B = -\frac{1}{4}}

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 8
    Last Post: March 22nd 2011, 05:57 PM
  2. Replies: 2
    Last Post: July 5th 2010, 09:48 PM
  3. Replies: 1
    Last Post: February 17th 2010, 04:58 PM
  4. Replies: 0
    Last Post: June 16th 2009, 01:43 PM
  5. Replies: 2
    Last Post: April 6th 2009, 09:57 PM

/mathhelpforum @mathhelpforum