<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] $$S=\frac{x+3}{2-x} = \frac{x-2+5}{2-x}=-1+\frac{5}{2-x}$$
$$S=-1+\frac 52\frac{1}{1-\dfrac x2}$$
$$S=-1+\frac 52\sum_{n=0}^{\infty}\left (\dfrac x 2\right )^n$$
Change the indice of the sum:
$$S=-1+\frac 52+\frac 52\sum_{n=1}^{\infty}\left (\dfrac x 2\right )^n$$
Finally:
$$S=\frac 32+\frac 52\sum_{n=1}^{\infty}\left (\dfrac x 2\right )^n$$
$$S=\frac 32+ 5\sum_{n=1}^{\infty}\dfrac {x^n}{ 2^{n+1}}$$
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