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!
On
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.
On
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:
$$ {\alpha \choose n} = \sum_{k=0}^n(-1)^k(k+1){\alpha+2 \choose n-k} \ \ (\forall \alpha \in \mathbb{C}, \ \forall \ n \in \mathbb{N}) $$
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*}
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*}