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?
On
Perhaps here is a way to start. We can define $$ f(x,y) = \sum_{j=0}^n (j+1)x^j (n-j+1) y^{n-j}, $$ where we ultimately may want to know $f(-1,2)$. Note this is very suggestive of differentiating a much simpler function. In other words, integrating wrt $x$ we get $$ I_x(x,y) = \sum_{j=0}^n x^{j+1} (n-j+1) y^{n-j} + C(y) $$ and integrating again wrt $y$ $$ I_{xy}(x,y) = \sum_{j=0}^n x^{j+1} y^{n-j+1} + \int C(y) dy + K(x). $$ If we let $C(y) = 0 = K(x)$ we have $I_{xy}(x,y)$ which should be easy to compute via straight geometric series. Then take mixed partial wrt $x$ and then $y$ (or the other way around), and evaluate at $x=-1,y=2$.
Perhaps a simpler way may be to note that $$ f(-1,2) = 2^n \sum_{j=0}^n (j+1) (n-j+1) (-2)^{-j} = A \sum_{j=0}^n 2^{-j} + B \sum_{j=0}^n j 2^{-j} + C \sum_{j=0}^n j^2 2^{-j}, $$ where you can derive $A,B,C$ by expanding the linear term product and simplifying, and the 3 sums are geometric series $\sum_k a^k$ and 2 its derivatives.
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}