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Math Help - [SOLVED] Partial fraction?

  1. #1
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    [SOLVED] Partial fraction?

    Find the partial fractions of

    \frac{r}{(2r - 1)(2r + 1)(2r + 3)}

    Thanks.
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  2. #2
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    Hello, Mark!

    Exactly where is your difficulty?


    Find the partial fractions of: . \frac{r}{(2r - 1)(2r + 1)(2r + 3)}

    We have: . \frac{r}{(2r-1)(2r+1)(2r+3)} \;=\;\frac{A}{2r-1} +\frac{B}{2r+1} + \frac{C}{2r+3}

    Then: . r \;=\;A(2r+1)(2r+3) + B(2r-1)(2r+3) + C(2r-1)(2r+1)


    \begin{array}{ccccccc}\text{Let }r = \tfrac{1}{2}\!: & \tfrac{1}{2}  &=& A(2)(4) + B(0) + C(0) & \Rightarrow & A \:=\:\frac{1}{16} \\ \\<br />
\text{Let }r = \text{-}\tfrac{1}{2}\!: & \text{-}\tfrac{1}{2} &=& A(0) + B(\text{-}2)(2) + C(0) & \Rightarrow & B \:=\:\frac{1}{8} \end{array}

    \begin{array}{ccccccc}\text{Let }r = \text{-}\tfrac{3}{2}\!: & \text{-}\tfrac{3}{2} &=& A(0) + B(0) + C(\text{-}4)(\text{-}2) & \Rightarrow & C \:=\:\text{-}\frac{3}{16}\end{array}


    Therefore: . \frac{r}{(2r-1)(2r+1)(2r+3)} \;=\;\frac{\frac{1}{16}}{2r-1} + \frac{\frac{1}{8}}{2r+1} - \frac{\frac{3}{16}}{2r+3}

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  3. #3
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    For A, I got \frac{1}{20}. Looks like it was a careless mistake done by me. Thanks, dude!
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