<s>[INST] Taylor Series and Differentiation with Sigma notation $f(x) = \frac{x}{(2-3x)^2}$ Use Term By Term Differentiation to Find the Taylor Series about $x$=3 for 
Give The Open Interval of Convergence and express as sigma notation 
 $\sum A_n(x-3)^n$
$f(x) = \frac{x}{(2-3x)^2}$
So I have Found the Taylor Series for 1/(2-3x) to be 
$\sum{(-3)^{n}\cdot(x-3)^{n} }\cdot{(-7)^ {n+1}}$
How Do you find the original function taylor series and its interval of convergence and then express in sigma notation of the form  $\sum A_n(x-3)^n$
 [/INST] you can make a change of variable $x = 3 + h.$  then we have 
$$\begin{align} f(x) &= \frac{x}{(2-3x)^2} = \frac{3+h}{(2 - 9-3h)^2}=\frac 1{49}(3+h)\left(1 + \frac{3h}7\right)^{-2} \\
&=\frac 1{49}(3+h)\left(1 -\frac 21 \frac{3h}7 + \frac{2\cdot3}{1\cdot2}\left(\frac{3h}{7}\right)^2 - \frac{2\cdot3 \cdot 4}{1\cdot2\cdot 3}\left(\frac{3h}{7}\right)^2 + \ldots\right)   \end{align} $$
i will let you multiply out the two factors and collect the like terms.
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