Write (n+2)/(2n+1) in the form (1/A)+(B/2n+1) where A and B are constants to be found.
I have that the answer is ((1/2)*(2n+1)+(3/2))/2n+1 and that equals to 1/2+3/(2*(2n+1)
I dont understand how my teacher did to get that answer, can someone help me?
There are several ways to go about this. The way I would go is probably the following: Get the numerator to look like the denominator plus something. The denominator has a 2n in it and the numerator has a 1n so we need to multiply the numerator by 2 and, in order to keep the expression the same, divide by 2
\(\displaystyle \frac{n+2}{2n+1}\, =\, \frac{1}{2} \frac{2(n+2)}{2n+1}\, =\, \frac{1}{2} \frac{2n+4}{2n+1}\)
I only need 1 part of that 4 in the numerator so
\(\displaystyle \frac{n+2}{2n+1}\, =\, \frac{1}{2} \frac{(2n+1) + 3}{2n+1}\)
Now divide each term by the denominator
\(\displaystyle \frac{n+2}{2n+1}\, =\, \frac{1}{2} [\frac{2n+1}{2n+1}\, +\, \frac{3}{2n+1}]\, =\, \frac{1}{2} [1\, +\, 3 \frac{1}{2n+1}]\)
and distribute the 1/2
\(\displaystyle \frac{n+2}{2n+1}\, =\, \frac{1}{2}\, +\, \frac{3}{2} \frac{1}{2n+1}\)
Since the equation must be true for all n, another way which works is to substitute two 'nice' particular values for n to get two equations in two unknowns [the 1/A and B]. If we use 0 and -2 for n we get
1/A + B = 2
1/A - (1/3) B = 0
and we can solve for A and B.
