For this problem, I must simplify the fraction so I end up with a sum of two different fractions:
$$1 - \frac{3}{e^n}$$ and $$2 - \frac{1}{m(m+1)}$$
and ultimately end up in a sort of sum between them. I believe I have no problem getting to the second one, but whenever I try to get to the first one, I end not being able to get rid of the $(m^2 + m)$ in the numerator and/or denominator. Furthermore, when I sum up the two fractions, I don't end up getting to the beginning fraction.
This is how I've attempted to do it:
$$\frac{2e^nm^2 + 2e^nm - e^n -6m^2 - 6m +3}{e^nm^2 + e^nm} = \frac{e^n(2m^2 + 2m -1) -6(m^2 +m)+3}{e^n(m^2+m)} = \frac{2m^2+2m-1}{m^2 + m} + \frac{-6(m^2+m)+3}{e^n(m^2+m)}$$ You can see that the first part of my solution ultimately comes out to the second faction listed above, but the whatever I try to do with the second part of my solution, I nothing works towards the end goal.
I would be beyond thankful if someone could explain how to get from start to finish as simply cannot wrap my head around it.
Expected end result: $$(1 - \frac{3}{e^n}) + (2 - \frac{1}{m(m+1)})$$
$$\frac{2e^nm^2 + 2e^nm - e^n -6m^2 - 6m +3}{e^nm^2 + e^nm}\\ =\frac{e^n(2m^2 + 2m - 1) -6(m^2 + m) +3}{e^nm^2 + e^nm}\\ =\frac{2m^2 + 2m - 1}{m^2 + m}-\frac{6}{e^n}+ \frac{3}{e^nm^2 + e^nm}\\ =2-\frac{1}{m^2 + m}-\frac{6}{e^n}+ \frac{3}{e^nm^2 + e^nm}\\=\frac{3}{me^{n}}-\frac{3}{e^{n}\left(m+1\right)}-\frac{6}{e^{n}}-\frac{1}{m}+\frac{1}{m+1}+2$$
The expected end result doesn't match your original fraction