# [SOLVED] Partial fraction?

• Aug 10th 2009, 06:50 AM
mark1950
[SOLVED] Partial fraction?
Find the partial fractions of

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

Thanks.
• Aug 10th 2009, 07:27 AM
Soroban
Hello, Mark!

Quote:

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

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

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

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

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

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

• Aug 10th 2009, 02:28 PM
mark1950
For A, I got $\displaystyle \frac{1}{20}$. :p Looks like it was a careless mistake done by me. Thanks, dude!