partial fraction

**saiyanmx89** · March 29th 2009, 07:41 AM

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

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

**running-gag** · March 29th 2009, 08:00 AM

Originally Posted by saiyanmx89

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

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

No

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

$\frac {1}{(2n+1)(2n+3)} = \frac {A}{2n+1} + \frac {B}{2n+3}$
Use the same denominator to find A and B

**skeeter** · March 29th 2009, 08:04 AM

Originally Posted by saiyanmx89

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

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

$\frac {1}{(2n+1)(2n+3)} = \frac{A}{2n+1} + \frac{B}{2n+3}$

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

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

$1 = -2B$

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

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

$1 = 2A$

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

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

**saiyanmx89** · March 29th 2009, 08:05 AM

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.

**running-gag** · March 29th 2009, 08:09 AM

The same denominator is $(2n+1)(2n+3)$

Thread: partial fraction

LinkBack

Thread Tools

Display

partial fraction

Similar Math Help Forum Discussions

Help for Partial Fraction

Partial Fraction Help?

please help improper fraction into partial fraction

partial fraction from improper fraction

Partial fraction

Search Tags