<s>[INST] 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?
 [/INST] $\begin{align}\frac{3 + x}{2 - x} &=\frac{-x-3}{x-2}\end{align}$
Use long division to find the quotient and remainder of $\frac{-x-3}{x-2}$ and then rewrite it as the quotient plus the remainder over the denominator:   
$\begin{align}\frac{-x-3}{x-2} 
&= -1-\frac{5}{x-2}=-1+\frac{5}{2-x}\\
&=-1-\frac{\frac52}{1-\frac x2}\\
&=-1+\frac{5}{2}.\frac{1}{1-\frac{x}{2}} \end{align}$
Since,
$\frac{1}{1-t} = 1+t+t^2+t^3+\>...$
$\begin{align}-1+\frac{5}{2}.\frac{1}{1-\frac{x}{2}}
&=-1+\frac52\left(1+ \frac x2 + \frac {x^2}{2^2} + \frac {x^3}{2^3}+\> …\right)\\
&=-1+\frac52+\frac52\left(\frac x2 + \frac {x^2}{2^2} + \frac {x^3}{2^3}+\> …\right)\\
&=\frac32+\frac52 \sum_{n=1}^∞ \frac{x^n}{ 2^{n}}\\
&=\frac32+\sum_{n=1}^∞ \frac{5x^n}{ 2^{n+1}}\end{align}$
Here is the answer that same with your professor. Have a nice day :D
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