Finding the MacLaurin series of $\frac{x+3}{2-x}$ I did:
$$\frac{x+3}{2-x} = \frac{x}{2-x}+\frac{3}{2-x} = x(\frac{1}{2-x})+3(\frac{1}{2-x}) = \\
= \frac{x}{2}(\frac{1}{1-\frac{x}{2}})+\frac{3}{2}(\frac{1}{1-\frac{x}{2}}) = \\
= \frac{x+3}{2}\sum(\frac{x}{2})^n = ...?$$
The answer my professor got was $\frac{(3 + x)}{(2 - x)} = \frac{3}{2} + \sum_{n=1}^∞ \frac{x^n5}{ 2^{n+1}}$
Unfortunately I forgot to write down how he did it. How do I get that solution and is mine necessarily incorrect/incomplete?

Use $\frac{1}{1-t} = 1+t+t^2+t^3+\>...$ to get,
$$\frac{3 + x}{2 - x} = -1 + \frac{\frac52}{1 - \frac x2} 
= -1+\frac52\left(1+ \frac x2 + \frac {x^2}{2^2} + \frac {x^3}{2^3}+\> …\right)= \frac{3}{2} + \sum_{n=1}^∞ \frac{5x^n}{ 2^{n+1}} $$ 
