I am going through some old Calculus-tasks in preparation for an upcoming exam, and a seemingly simple task is being stubborn with me. We are simply to find the power-series of the function $$\frac{1}{(2-x)^2}$$
Now, we have our basic power-series:
$$\frac{1}{1-x} = 1 + x + x^2 + x^3 + \cdots$$
So we have
$$\frac{1}{2-x} = \frac{1}{2(1-\frac{1}{2}x)} = \frac{\frac{1}{2}}{1-\frac{1}{2}x} = \frac{1}{2} + \frac{1}{4}x + \frac{1}{8}x^2 + \frac{1}{16}x^3 + \cdots$$
Let $f(x) = \frac{1}{2-x}$. Using the chain-rule, we set $u = 2-x$ and $g(u) = \frac{1}{u}$. Then we have $f'(x) = g'(u) * u' = -\frac{1}{u^2}*(-1) = \frac{1}{(2-x)^2}$
So we have $$\frac{1}{(2-x)^2} = \frac{1}{4} + \frac{1}{4}x + \frac{3}{16}x^2 \cdots$$
The text does not agree on this answer. In particular, it claims that the first term should be $\frac{1}{2}$
Using Newton's generalised binomial theorem or the Binomial Series and assuming the convergence ,
$$\frac1{(2-x)^2} =\frac14\left(1-\frac x2\right)^{-2}$$ $$=\frac14\left(1+\frac x2\cdot2+\frac{-2(-2-1)}{2!}\left(\frac x2\right)^2+\cdots\right)=\cdots$$