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