What are the steps for deriving a complicated generalization of a partial sum of a taylor series? I looked at the Taylor series for $$-\frac{x}{x-2}$$ and found it to be $$ \sum_{k=1}^{\infty}\frac{x^k}{2^k}$$
but then I also found that this series' partial sum is a bit more complicated in the form of
$$\frac{x 2^{-k}(2^{k}+x^{k})}{x+2} $$
My question is: how was such a partial sum derived, and how would you derive it for something more complicated, like for instance
$$\sum_{k=1}^{\infty}\frac{x^{k^2}}{k!}?$$
 A: Consider that for any $r\neq 1$, $$\sum_{k=1}^n r^k = \frac{r(1-r^n)}{1-r}$$ Therefore, we can just take $r = x/2$ (for $x\neq 2$) to get the correct partial sum, which is $$\sum_{k=1}^n \frac{x^k}{2^k} = \frac{(x/2)(1-(x/2)^n)}{1-x/2} = \frac{x(2^n-x^n)}{2^n(2-x)}$$ I don't think that there's a closed form for the second sum you've mentioned, though.
A: The Taylor series has a strict formula. Suppose $f(x)$ has any derivative at $x_0$, then
$$f(x)=\sum _{n=0}^{\infty } \dfrac{f^{(n)}(x_0)}{n!}(x-x_0)^n$$
where $f^{(n)}$ are the derivatives of the function, with the convention that $f^{(0)}=f,\;f^{(1)}=f',\ldots$
If $x_0=0$ then the formula simplifies in the MacLaurin version
$$f(x)=\sum _{n=0}^{\infty } \dfrac{f^{(n)}(0)}{n!}x^n\quad(*)$$
You have to compute all derivatives at $x=0$ and plug them in the formula.
In your example $f(x)=-\dfrac{x}{x-2}$ the first derivatives from first to sixth are
$$\frac{2}{(x-2)^2},-\frac{4}{(x-2)^3},\frac{12}{(x-2)^4},-\frac{48}{(x-2)^5},\frac{240}{(x-2)^6},-\frac{1440}{(x-2)^7}$$
If you plug $x=0$ you get
$$f^{1}(0)=\frac{1}{2},f^{2}(0)=\frac{1}{2},f^{3}(0)=\frac{3}{4},f^{4}(0)=\frac{3}{2},f^{5}(0)=\frac{15}{4},f^{6}(0)=\frac{45}{4}$$
and now plug in the MacLaurin formula $(*)$
$$-\dfrac{x}{x-2}=f(0)+f^{1}(0)x+\frac{f^{2}(0)}{2!}x^2+\frac{f^{3}(0)}{3!}x^3+\frac{f^{4}(0)}{4!}x^4+\frac{f^{5}(0)}{5!}x^5+\frac{f^{6}(0)}{6!}x^6+\ldots$$
and then
$$-\dfrac{x}{x-2}=\frac{1}{2}x+\frac{\frac{1}{2}}{2!}x^2+\frac{\frac{3}{2}}{4!}x^3+\frac{\frac{3}{2}}{4!}x^4+\frac{\frac{15}{4}}{5!}x^5+\frac{\frac{45}{4}}{6!}x^6+\ldots$$
and finally
$$-\dfrac{x}{x-2}=\frac{1}{2}x+\frac{1}{4}x^2+\frac{1}{8}x^3+\frac{1}{16}x^4+\frac{1}{32}x^5+\frac{1}{64}x^6+\ldots$$
It's not a "choice", is the formula 
Hope it helps
A: In this particular case, the partial sum is easy to find:
$$f(x)=\sum_{k=1}^\infty\frac{x^k}{2^k}\implies \frac{x^n}{2^n}f(x)=\frac{x^n}{2^n}\sum_{k=1}^\infty\frac{x^k}{2^k}=\sum_{k=1}^\infty\frac{x^{k+n}}{2^{k+n}}=\sum_{k=n+1}^\infty\frac{x^k}{2^k}.$$
Then by subtraction,
$$\sum_{k=1}^n\frac{x^k}{2^k}=f(x)\left(1-\frac{x^k}{2^k}\right).$$
