Taylor series about 1 I was trying to solve the Taylor series about one for $\dfrac{x^2}{2 - x}$ but my answer seems to be wrong. I got $T(x) =  \frac{1}{n!}(x-1)^n$.

We are looking for a representation
\begin{align*}
\sum_{n=0}^\infty a_n(x-1)^n
\end{align*}

We obtain
\begin{align*}
\color{blue}{\frac{x^2}{2-x}}&=\frac{(x-1+1)^2}{1-(x-1)}\\
&=\left((x-1)^2+2(x-1)+1\right)\sum_{n=0}^\infty (x-1)^n\\
&=\sum_{n=0}^\infty(x-1)^{n+2}+2\sum_{n=0}^\infty(x-1)^{n+1}+\sum_{n=0}^\infty(x-1)^n\\
&=\sum_{n=2}^\infty(x-1)^{n}+2\sum_{n=1}^\infty(x-1)^{n}+\sum_{n=0}^\infty(x-1)^n\\
&\,\,\color{blue}{=1+3(x-1)+4\sum_{n=2}^\infty(x-1)^n}
\end{align*}

