Find A and B so that 1 / [(4n-3)(4n+1)] = A / (4n-3) + B / (4n+1).
Thanks for your time, efforts, and help!
You have,
$\displaystyle \frac{1}{(4n-3)(4n+1)}=\frac{A}{4n-3}+\frac{B}{4n+1}$
Multiply through by,
$\displaystyle (4n-3)(4n+1)$,
$\displaystyle 1=A(4n+1)+B(4n-3)$
$\displaystyle 1=n(4A+4B)+(A-3B)$
We have,
$\displaystyle 4A+4B=0$
$\displaystyle A-3B=1$
Solve, to get,
$\displaystyle A=1/4,B=-1/4$
Here is one way.
1 /[(4n-3)(4n+1)] = A/(4n-3) +B/(4n+1)
Multiply both sides by (4n-3)(4n+1)
1 = A(4n+1) +B(4n-3) ---------------(i)
So that B is eliminated, make (4n-3) equal to zero.
4n -3 = 0
n = 3/4
So, when n=3/4, in (i),
1 = A(4(3/4) +1) +B(4(3/4) -3)
1 = A(3+1) +B(3-3)
1 = 4A
A = 1/4 ----------------------answer.
So that A is eliminated, make (4n+1) equal to zero.
4n +1 = 0
n = -1/4
So, when n = -1/4, in (i),
1 = A(4(-1/4) +1) +B(4(-1/4) -3)
1 = A(-1 +1) +B(-1 -3)
1 = -4B
B = -1/4 ----------------------answer.
Hello, Mr_Green!
Another method . . .
Multiply through by $\displaystyle (4n-3)(4n+1)\!:\;\;1 \;=\;(4n+1)A + (4n-3)B$Find $\displaystyle A$ and $\displaystyle B$ so that: .$\displaystyle \frac{1}{(4n-3)(4n+1)} \:= \:\frac{A}{4n-3} + \frac{B}{4n+1}$
Let $\displaystyle n = \frac{3}{4}\!:\;\;1 \:=\:\left[4\left(\frac{3}{4}\right) + 1\right]A + \left[4\left(\frac{3}{4}\right) - 3\right] B$
. . $\displaystyle 1 \:=\:4\!\cdot\!A + 0\!\cdot\!B\quad\Rightarrow\quad\boxed{ A\,=\,\frac{1}{4}}$
Let $\displaystyle n = -\frac{1}{4}\!:\;\;1\:=\:\left[4\left(-\frac{1}{4}\right) + 1\right]A + \left[4\left(-\frac{1}{4}\right) - 3\right]B$
. . $\displaystyle 1 \:=\:0\!\cdot\!A - 4\!\cdot\!B\quad\Rightarrow\quad\boxed{ B = -\frac{1}{4}}$