Taylor polynomial of $\frac{1}{2-x}$ Can someone show how to find the Taylor polynomial of $\frac{1}{2-x}$?
I tried this: $\frac{1}{2-x}=\frac{1}{1-(x-1)}$ and then use that $ \ T_n(\frac{1}{1-x})=1+x+\dots +x^n.$ But this gives $1+(x-1)+\dots +(x-1)^n$ which is the wrong answer. Why?

We want to find $a_0, a_1, a_2,\ldots$ so that  $$\frac1{2-x} = a_0 + a_1x + a_2x^2 + \ldots.$$
Multiply both sides by $2-x$: $$
\begin{align}
1 & = (2-x)a_0 + (2-x)a_1x + (2-x)a_2x^2 + \cdots\\
& = (2a_0-a_0x)+(2a_1x-a_1x^2) + (2a_2x^2 - a_2x^3) + \cdots \\
\end{align}$$
Combine like terms:
$$ 1 = 2a_0 + (2a_1-a_0)x + (2a_2 -a_1)x^2 + \cdots$$
Equate coefficients on both sides:
$$\begin{align}
2a_0 & = 1 \\
2a_1 - a_0 & = 0 \\
2a_2 - a_1 & = 0 \\
& \vdots
\end{align}$$
Now just solve the equations starting with the top one:
$$\begin{align}
a_0 &= \frac12 \\
a_1 &= \frac14\\
a_2 &= \frac18 \\
& \vdots
\end{align}$$
