Write out first four terms of Taylor series $f(x)= \frac{1}{(2-x)^2}$

Question

To do it the simple way, I know $f(x)=\frac{1}{(2-x)^2}$ can be written as $$g(x)= \frac{1}{2-x}$$ and then you can differentiate term by term. What exactly does that mean?

Answer 1

You would write $g(x)=a_0+a_1x+a_2x^2+a_3x^3+\ldots$.
Then $g'(x)=a_1+2a_2x+3a_3x^2+\ldots$
Differentiate term by term is just what it says.

Answer 2

$$ g(x) = (2-x)^{-1}\\ g'(x) = (2-x)^{-2} (-1)\\ = - f(x)\\ f(x) = -g'(x) $$

$$ g(x) = \frac{1}{2} \frac{1}{1-x/2}\\ = \frac{1}{2} \sum_{r=0}^\infty (\frac{x}{2})^r\\ -g'(x) = - \frac{1}{2} \sum_{r=0}^\infty \frac{rx^{r-1}}{2^r} $$