Show that the following expansion is valid for first $n$ terms The question is as follows, I've gone ahead and attempted as much as I could of it twice, however the final answer that I'm receiving does not match.

I'm omitting the partial fractions steps; if they are required I'll gladly put them up.
$$\frac{3x^2+2x}{x^3-2x^2+3x-6} \equiv \frac{5}{7(x-2)} + \frac{5x-4}{7(x^2+3)}$$
Thus for what range of values of $x$ is the expansion valid?
$$-\frac{1}{3}x - \frac{2}{3}x^2 - \frac{2}{9}x^3+...$$
In order to bring the terms as unity (pardon is I'm using the wrong terminology) I did the following $$7(x-2) = 14(\frac{1}{2}x-1) $$ $$\Rightarrow 7(x^2+3) = 3(\frac{1}{3}x+1)^{\frac{1}{2}}$$
And expanded accordingly using $1 + nx + \frac{n(n-1)(x^2)}{2!} + \frac{n(n-1)(n-2)(x^3)}{3!} + ...$
And my expansions boiled down equivalently to $$(14*3)(1-\frac{1}{2}x+\frac{1}{4}x^2-\frac{1}{8}x^3)(1+\frac{1}{6}x-\frac{1}{72}x^2+\frac{1}{432}x^3)$$
However seeing this last bit, I realised that I must've messed up immensely somewhere. 
Essentially I wanted to boil down the expansion to the one above and determine the values for which $-1 < x < 1$ in order for the expansion to be valid.
 A: You seem to have the wrong partial fractions. I get
\begin{eqnarray*}
\frac{3x^2+2x}{x^3-2x^2+3x-6} &=& \frac{16}{7(x-2)} + \frac{5x
+24}{7(x^2+3)} \\
&=& \frac{-8}{7} \frac{1}{1-\frac{x}{2}} + \frac{5x+24}{21} \frac{1}{1-\frac{-x^2}{3}}. \\
&& \\
\text{Using the series for $\;\frac{1}{1-x}$} && \text{for both of the above fractions} : \\
\frac{-8}{7} \frac{1}{1-\frac{x}{2}} &=& \frac{-8}{7} \left(1 + \frac{x}{2} + \left(\frac{x}{2}\right)^2 + \left(\frac{x}{2}\right)^3 + \cdots \right)\qquad\text{for $|x| \lt 2$} . \\
&& \\
\frac{5x+24}{21} \frac{1}{1-\frac{-x^2}{3}} &=& \frac{5x+24}{21} \left(1 - \frac{x^2}{3} + \left(\frac{x^2}{3}\right)^2 - \left(\frac{x^2}{3}\right)^3 + \cdots \right) \\
&=& \frac{1}{21} \left(24 + 5x - 8x^2 - \frac{5x^3}{3} + \cdots \right)\qquad\text{for $|x| \lt \sqrt{3}$} . \\
&& \\
\text{Summing,}\quad \frac{3x^2+2x}{x^3-2x^2+3x-6} &=& -\frac{1}{3}x - \frac{2}{3}x^2 - \frac{2}{9}x^3+\cdots . \\
&& \\
\end{eqnarray*}
From the radii of convergence of the partial fractions, this series is valid for $|x|\lt \sqrt{3}$.
