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.
$$F(x)=\frac{(1-2x+2x^2)}{(1-x)^2}$$
 A: 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.
A: HINT
Note that if $G(x) = (1-x)^{-1}$ then $G'(x) = (1-x)^{-2}$, and you have
$$
G(x)  = \sum_{k=0}^\infty x^k\\
G'(x) = \sum_{k=1}^\infty kx^{k-1}\\
$$
and
$$
F(x) = G'(x) (1-2x+2x^2)
$$
Can you finish this?
UPDATE
$$
\begin{split}
\left[x^{25}\right] F(x)
 &= \left[x^{25}\right] (1-2x+2x^2) \sum_{k=1}^\infty kx^{k-1} \\
 &= \left[x^{25}\right] \sum_{k=1}^\infty kx^{k-1}
    -2 \left[x^{24}\right] \sum_{k=1}^\infty kx^{k-1}
    +2 \left[x^{23}\right] \sum_{k=1}^\infty kx^{k-1}\\
\end{split}
$$
Can you finish this?
A: First simplify the rational function as follows:
$$
\frac{1-2x+2x^2}{(1-x)^2}
=\frac{(1-x)^2+x^2}{(1-x)^2}
=1+\frac{x^2}{(1-x)^2}.\tag{1}
$$
Then use the identity 
$$
\frac{1}{(1-x)^k}=\sum_{n=0}^\infty \binom{k+n-1}{k-1}x^n\tag{2}
$$
(which can be obtained by repeatedly differentiating the geometric series) where $k\geq 1$ to get that
$$
[x^n]\left(\frac{1}{(1-x)^2}\right)=n+1\tag{3}
$$
and hence
$$
[x^n]\left(\frac{x^2}{(1-x)^2}\right)=n-1.\tag{4}
$$
Finally, to answer the question note that
$$
[x^{25}]\left(\frac{1-2x+2x^2}{(1-x)^2}\right)=
[x^{25}]\left(1+\frac{x^2}{(1-x)^2}\right)=[x^{25}]\left(\frac{x^2}{(1-x)^2}\right)=25-1=\color{blue}{24}
$$
by (1) and (4).
