1. ## partial fraction

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

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

2. Originally Posted by saiyanmx89
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

3. Originally Posted by saiyanmx89
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 $\displaystyle n = -\frac{1}{2}$ ...

$\displaystyle 1 = 2A$

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

$\displaystyle \frac {1}{(2n+1)(2n+3)} = \frac{1}{2}\left(\frac{1}{2n+1} - \frac{1}{2n+3}\right)$

4. 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.

5. The same denominator is $\displaystyle (2n+1)(2n+3)$