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?
 A: Use $\frac{1}{1-t} = 1+t+t^2+t^3+\>...$ to get,
$$\frac{3 + x}{2 - x} = -1 + \frac{\frac52}{1 - \frac x2} 
= -1+\frac52\left(1+ \frac x2 + \frac {x^2}{2^2} + \frac {x^3}{2^3}+\> …\right)= \frac{3}{2} + \sum_{n=1}^∞ \frac{5x^n}{ 2^{n+1}} $$ 
A: \begin{align}
\frac{x+3}2\sum_{n=0}^\infty\left(\frac{x}2\right)^n
&=\frac{x}2\sum_{n=0}^\infty\frac{x^n}{2^n}+\frac32\sum_{n=0}^\infty\frac{x^n}{2^n}\\
&=\sum_{n=0}^\infty\frac{x^{n+1}}{2^{n+1}}+\sum_{n=0}^\infty\frac{3x^n}{2^{n+1}}\\
&=\overbrace{\sum_{n=1}^\infty\frac{x^n}{2^n}}^{n+1\,\mapsto\, n}+\overbrace{\left(\frac32+\sum_{n=1}^\infty\frac{3x^n}{2^{n+1}}\right)}^{\text{remove constant term }(n=0)}\\
&=\frac32+\sum_{n=1}^\infty\left(\frac{x^n}{2^n}+\frac{3x^n}{2^{n+1}}\right)\\
&=\frac32+\sum_{n=1}^\infty\frac{5x^n}{2^{n+1}}\\
\end{align}
A: $$S=\frac{x+3}{2-x} = \frac{x-2+5}{2-x}=-1+\frac{5}{2-x}$$
$$S=-1+\frac 52\frac{1}{1-\dfrac x2}$$
$$S=-1+\frac 52\sum_{n=0}^{\infty}\left (\dfrac x 2\right )^n$$
Change the indice of the sum:
$$S=-1+\frac 52+\frac 52\sum_{n=1}^{\infty}\left (\dfrac x 2\right )^n$$
Finally:
$$S=\frac 32+\frac 52\sum_{n=1}^{\infty}\left (\dfrac x 2\right )^n$$
$$S=\frac 32+ 5\sum_{n=1}^{\infty}\dfrac {x^n}{ 2^{n+1}}$$
A: $\begin{align}\frac{3 + x}{2 - x} &=\frac{-x-3}{x-2}\end{align}$
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
