Q:

obtaining a simplified expression for the coefficient of $x^n$

I was trying to find the coefficient of $x^n$ in the expansion of $(1+x)^{-2}(1-2x)^{-2},$ denoted $[x^n]\{(1+x)^{-2}(1-2x)^{-2}\}$. Using the negative binomial theorem, I know that it is equal to
$$
\begin{split}
\sum_{j=0}^n &([x^j](1+x)^{-2})([x^{n-j}](1-2x)^{-2}) \\
  &= \sum_{j=0}^n {j+1\choose 1}(-1)^j{n-j+1\choose 1}2^{n-j} \\
  &= \sum_{j=0}^n (j+1)(n-j+1)(-1)^j2^{n-j}.
\end{split}
$$

However, I was wondering if there was a way to simplify this expression even further?

A:

As suggested by @AnginaSeng, you can apply partial fraction decomposition:
\begin{align}
\frac{1}{(1+x)^2(1-2x)^2}
&=\frac{1/9}{(1+x)^2}+\frac{4/27}{1+x}+\frac{4/9}{(1-2x)^2}+\frac{8/27}{1-2x}\\
&=\frac{1}{9}\sum_{n \ge 0}\binom{n+1}{1}(-x)^n+\frac{4}{27}\sum_{n\ge 0} (-x)^n+\frac{4}{9}\sum_{n \ge 0} \binom{n+1}{1}(2x)^n+\frac{8}{27}\sum_{n\ge 0} (2x)^n\\
&=\sum_{n \ge 0}\left(\frac{1}{9}\binom{n+1}{1}(-1)^n+\frac{4}{27}(-1)^n+\frac{4}{9}\binom{n+1}{1}2^n+\frac{8}{27} 2^n\right) x^n\\
&=\sum_{n \ge 0}\left(\color{blue}{\frac{(3n+7)(-1)^n+(12n+20)2^n}{27}}\right) x^n
\end{align}

