Generating function and binomial coefficient a) I have to find and expression for sequence of $b_n$ in terms of generating functions of the sequence of $a_n$
$$b_n = (-1)^{n}(n+1)a_0 +(-1)^{n-1}n a_1+...+(-1)2a_{n-1}+a_n$$ with $$a_n = a_{n-1} +8a_{n-2} -12a_{n-3} +25(-3)^{n-2} + 32n^2 -64$$
b) I have to use the result of a) to prove this identity
$${\beta \choose n} = \sum_{x = 0}^{n}(-1)^{x}(x+1) {\beta+2 \choose {n-x}}$$ with $\beta$ is a complex number
Please shed some lights. Any help or hints would be greatly appreciated
 A: 
Case a.) We consider a generating function $A(z)=\sum_{n=0}^\infty a_n z^n$ and are looking for a generating function $B(z)=\sum_{n=0}^\infty b_n z^n$ with
  \begin{align*}
b_n&=(-1)^n(n+1)a_0+(-1)^{n-1}na_1+\cdots -2a_{n-1}+a_n\\
&=\sum_{k=0}^n(-1)^{n-k}(n+1-k) a_k\tag{1}
\end{align*}
We apply the Cauchy product formula. It is convenient to use the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ in a series. This way we can write for instance
  \begin{align*}
[z^n]A(z)=a_n
\end{align*}
We obtain from (1):
  \begin{align*}
\color{blue}{b_n}
&=\sum_{k=0}^n(-1)^{n-k}(n+1-k) a_k\\
&=\sum_{k=0}^n(-1)^{k}(k+1) a_{n-k}\tag{2}\\
&=\sum_{k=0}^n\left([z^k]\frac{1}{(1+z)^2}\right)\left([z^{n-k}]A(z)\right)\tag{3}\\
&\,\,\color{blue}{=[z^n]\frac{1}{(1+z)^{2}}A(z)}
\end{align*}
  and conclude the relation (1) in terms of generating functions is given as
  \begin{align*}
\color{blue}{B(z)=\frac{A(z)}{(1+z)^2}}\tag{4}
\end{align*}

Comment:


*

*In (2) we change the order of summation by replacing the index $k$ with $n-k$.

*In (3) we use the series representation
\begin{align*}
\sum_{k=0}^\infty(-1)^k(k+1)z^k=\frac{1}{(1+z)^2}
\end{align*}
and apply the Cauchy product formula.

Case b.) In order to show the binomial identity for $\beta\in\mathbb{C}$ and $n\in\mathbb{N}$
  \begin{align*}
\binom{\beta}{n}=\sum_{k=0}^n(-1)^k(k+1)\binom{\beta+2}{n-k}\tag{5}
\end{align*}
we just need the relationsship  between $b_n$ and $a_n$ as stated in (1).
  Here we are in a  situation as shown in (2) with $a_n=b_n$, namely
  \begin{align*}
a_n=\sum_{k=0}^n(-1)^{k}(k+1) a_{n-k}
\end{align*}
  This means that recalling the binomial series expansion and setting
  \begin{align*}
A(z)&=\sum_{k=0}^\infty\binom{\beta+2}{k}z^k=(1+z)^{\beta+2}\\
B(z)&=\sum_{k=0}^\infty\binom{\beta}{k}z^k=(1+z)^{\beta}\\
\end{align*}
  and we see the binomial identity (5) in terms of generating functions is
  \begin{align*}
\color{blue}{(1+z)^{\beta}=\frac{1}{(1+z)^2}(1+z)^{\beta+2}}
\end{align*}
  which proves the claim.

