Proving a complex binomial identity I would like to prove an identity:
$$\binom{\alpha}{n} = \sum_{k=0}^n(-1)^k(k+1)\binom{\alpha + 2}{n-k}$$
Where $\alpha$ is complex.
I have already found that if you have two sequences related by the identity $$b_n = \sum_{k=0}^n(-1)^{k}(1+k)a_{n-k}$$ you can write the generating function for $b_n$ (which I'll write as $B(z)$) in terms of the generating function for $a_n$ as follows:
$$B(z) = \frac{A(z)}{(1+z)^2}$$
How do I prove this identity, using the above fact?
Thanks in advance!
 A: 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](1+z)^\alpha=\binom{\alpha}{n}
\end{align*}

We obtain for $\alpha\in\mathbb{C}$ and $n\in\mathbb{N}$:
  \begin{align*}
\color{blue}{\sum_{k=0}^n}&\color{blue}{(-1)^k(k+1)\binom{\alpha+2}{n-k}}\\
&=\sum_{k=0}^n\left([z^k]\frac{1}{(1+z)^2}\right)\left([z^{n-k}](1+z)^{\alpha+2}\right)\tag{1}\\
&=[z^n](1+z)^{\alpha}\\
&\,\,\color{blue}{=\binom{\alpha}{n}}
\end{align*}
  and the claim follows.

Comment:


*

*In (1) we use the Cauchy product formula\begin{align*}
A(z)&=\sum_{k=0}^\infty a_kz^k,\qquad B(z)=\sum_{j=0}^\infty b_jz^j\\
A(t)B(t)&=\sum_{n=0}^\infty\left(\sum_{{k+j=n}\atop{k,j\geq 0}}a_kb_j\right)t^n=\sum_{n=0}^\infty \left(\sum_{k=0}^n a_k b_{n-k}\right)t^n\\
&=\sum_{n=0}^\infty\sum_{k=0}^n \left([z^k]A(z)\right)\left([z^{n-k}]B(z)\right)t^n
\end{align*}
In the case above we have
\begin{align*}
A(z)&=\sum_{k=0}^\infty (-1)^k(k+1)z^k=\frac{1}{(1+z)^2}\\
B(z)&=\sum_{k=0}^\infty \binom{\alpha+2}{k}z^k=(1+z)^{\alpha+2}
\end{align*}
A: The classic recursion for Pascal's triangle is
 $\;{\alpha\choose n}+{\alpha\choose n-1}={\alpha+1\choose n},\;$ applied twice giving $\;{\alpha+2\choose n}.\;$ Also
 $\;(1+z)B(z)=A(z)\;$ is equivalent to $\;a_n=b_n+b_{n-1},\;$ and applied twice  we are done.
A: Edit: this proof does not depend on generating functions, which I missed was a requirement on the original question. Since this is rather elementary, I figured it could be of use to post it anyway.
First, let's show that this holds for $m \in \mathbb{N}$. We can do induction on the following proposition: for each $m$, it holds that 
$$
{m \choose n} = \sum_{k=0}^n(-1)^k(k+1){m+2 \choose n-k}
$$
for any natural $n \leq m$. In effect, if $m = 1$, the only case to consider is 
$$
{1 \choose 1} = 1 = 3 - 2 = \sum_{k = 0}^1(-1)^k(k+1){3 \choose 1 - k}
$$
and if the proposition holds for $m$, 
$$
{m+1 \choose n} = {m \choose n} + {m \choose n-1} = \\ = \sum_{k=0}^n(-1)^k(k+1){m+2 \choose n-k} + \sum_{k=0}^{n-1}(-1)^k(k+1){m+2 \choose n-1-k} = \\ 
\sum_{k=0}^{n-1}(-1)^k(k+1)\big[{m+2 \choose n-k} + {m+2 \choose n-1-k}\big] + (-1)^n(n+1){m+2 \choose 0} = 
\\ = \sum_{k=0}^{n-1}(-1)^k(k+1){m+3 \choose n-k} + (-1)^n(n+1){m+2 \choose 0} = \\
= \sum_{k=0}^{n-1}(-1)^k(k+1){m+3 \choose n-k} + (-1)^n(n+1){m+3 \choose 0} = \sum_{k=0}^{n}(-1)^k(k+1){m+3 \choose n-k} 
$$
where here we're using the inductive hypothesis, that is, that the equality holds for $m$ and $n \leq  m$. To conclude the result, we're left with the case where $n = m+1$, and in this instance,
$$
\sum_{k=0}^{m+1}(-1)^k(k+1){m+3 \choose m+1-k} = \sum_{k=0}^{m+1}(-1)^k(k+1){m+3 \choose k+2} = \\ = \sum_{k=2}^{m+3}(-1)^k(k-1){m+3 \choose k} = \sum_{k=2}^{m+3}(-1)^kk{m+3 \choose k} - \sum_{k=2}^{m+3}(-1)^k{m+3 \choose k} = \\
= \big[{m+3 \choose 1}\big] - \big[-{m+3 \choose 0} + {m+3 \choose 1}\big] = 1 = {m+1 \choose m+1}
$$
Now we can prove the general case. Let's fix $n \in \mathbb{N}$, and define the following polynomials,
$$
P_n(\alpha) = {\alpha \choose n}, \ Q_n(\alpha) = \sum_{k=0}^n(-1)^k(k+1){\alpha+2 \choose n-k}
$$
with $P_n,Q_n \in \mathbb{C}[X]$. Since we've previously proved that $P_n \equiv Q_n$ for $\alpha \in \mathbb{N}_{\geq n}$, and these are complex valued polynomials in one variable, they must be the same. Moreover, since $n$ is arbitrary, we've proved that $P_n = Q_n$ for any $n \in \mathbb{N}$, which is precisely what we wanted:
$$
{\alpha \choose n} = \sum_{k=0}^n(-1)^k(k+1){\alpha+2 \choose n-k} \ \ (\forall \alpha \in \mathbb{C}, \ \forall \ n \in \mathbb{N})
$$
