Determining the coefficient of a generating function I've been doing some practice problems and I have encountered some questions. How would I determine the coefficient of $x^{25}$ in the generating function $F(x)$ with closed form expression? I can't factor the numerator which is why I did for the previous ones.
$$F(x)=\frac{(1-2x+2x^2)}{(1-x)^2}$$

It is convenient to use the coefficient of operator $[x^n]$ to denote the coefficient of $x^n$ in a series.

We obtain
  \begin{align*}
\color{blue}{[x^{25}]\frac{1-2x+2x^2}{(1-x)^2}}&=[x^{25}](1-2x+2x^2)\sum_{j=0}^\infty\binom{-2}{j}(-x)^j\tag{1}\\
&=\left([x^{25}]-2[x^{24}]+2[x^{23}]\right)\sum_{j=0}^\infty\binom{j+1}{1}x^j\tag{2}\\
&=26-2\cdot 25+2\cdot 24\tag{3}\\
&=\color{blue}{24}
\end{align*}

Comment:


*

*In (1) we apply the binomial series expansion.

*In (2) we use the linearity of the coefficient of operator and apply the rule
\begin{align*}
[x^{p-q}]A(x)=[x^p]x^qA(x)
\end{align*}
We also use the binomial identity
\begin{align*}
\binom{-p}{q}=\binom{p+q-1}{p-1}(-1)^q
\end{align*}

*In (3) we select the coefficients accordingly.
