Find A and B so that 1 / [(4n-3)(4n+1)] = A / (4n-3) + B / (4n+1).
Thanks for your time, efforts, and help!
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.