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Thread: partial fraction

  1. #1
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    partial fraction

    Did I write the correct partial fraction for this problem??

    $\displaystyle \frac {1}{(2n+1)(2n+3)}$ -> $\displaystyle \frac{3}{4} (\frac {1}{2n+1} + \frac {1}{2n+3})$
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  2. #2
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    Quote Originally Posted by saiyanmx89 View Post
    Did I write the correct partial fraction for this problem??

    $\displaystyle \frac {1}{(2n+1)(2n+3)}$ -> $\displaystyle \frac{3}{4} (\frac {1}{2n+1} + \frac {1}{2n+3})$
    No

    If you use the same denominator you will get $\displaystyle \frac{3}{4} \left(\frac {1}{2n+1} + \frac {1}{2n+3}\right) = \frac {3n+3}{(2n+1)(2n+3)}$

    $\displaystyle \frac {1}{(2n+1)(2n+3)} = \frac {A}{2n+1} + \frac {B}{2n+3}$
    Use the same denominator to find A and B
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  3. #3
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    Quote Originally Posted by saiyanmx89 View Post
    Did I write the correct partial fraction for this problem??

    $\displaystyle \frac {1}{(2n+1)(2n+3)}$ -> $\displaystyle \frac{3}{4} (\frac {1}{2n+1} + \frac {1}{2n+3})$
    $\displaystyle \frac {1}{(2n+1)(2n+3)} = \frac{A}{2n+1} + \frac{B}{2n+3}$

    $\displaystyle 1 = A(2n+3) + B(2n+1)$

    let $\displaystyle n = -\frac{3}{2}$ ...

    $\displaystyle 1 = -2B$

    $\displaystyle B = -\frac{1}{2}$

    let $\displaystyle n = -\frac{1}{2}$ ...

    $\displaystyle 1 = 2A$

    $\displaystyle A = \frac{1}{2}$

    $\displaystyle \frac {1}{(2n+1)(2n+3)} = \frac{1}{2}\left(\frac{1}{2n+1} - \frac{1}{2n+3}\right)$
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  4. #4
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    what do you mean by using the same denominator? you have (2n+1) and (2n+3). I found A and B to both be (3/4) but I know that is wrong. I don't understand what I did wrong.
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  5. #5
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    The same denominator is $\displaystyle (2n+1)(2n+3)$
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