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For $x\in\mathbb{R}$, $k\in\mathbb{N}$, $k\ge 1$, define $$\binom{x}{k}:=\prod_{j=1}^k\frac{x-j+1}{j}.$$

The claim is: For all $x,y\in\mathbb{R}$, $n\in\mathbb{N}$ such that $n\ge 1$, it is

$$ \binom{x+y}{n}=\sum\limits_{k=0}^n\binom{x}{n-k}\binom{y}{k}.$$

I think this requires an algebraic proof, and no proof by induction, or can I do both?

Trying it without induction I have the following problem: If I start with the right-hand site and replace the binomial coefficient by the product formula, I don't know how to proceed to go to the left-hand site, but I have found a similar formula Algebraic Proof that $\sum\limits_{k=0}^m \binom{r}{k} \binom{m+n-r}{m-k} = \binom{m+n}{m}$ Can you help me?

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It is convenient to use the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ in a series $A(z)$. This way we can write e.g. \begin{align*} \binom{x}{n}=[z^n](1+z)^x \end{align*}

We obtain \begin{align*} \color{blue}{\binom{x+y}{n}}&=[z^n](1+z)^{x+y}\\ &=[z^n]\left((1+z)^y(1+z)^x\right)\\ &=\sum_{k=0}^n\left([z^k](1+z)^y\right)\left([z^{n-k}](1+z)^x\right)\\ &\color{blue}{=\sum_{k=0}^n\binom{y}{k}\binom{x}{n-k}} \end{align*} and the claim follows.

Hint: This binomial identity is known as Chu-Vandermonde identity.

Markus Scheuer
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  • Thank you. Is it okay to prove it without using induction? –  Oct 16 '17 at 16:55
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    @induction: Yes, of course. As long as a proof using induction is not explicitly required I consider other types of proofs even advantageous. They often provide additional structural information about the problem. In this answer e.g. we see the connection with the Cauchy product of polynomials resp. series. – Markus Scheuer Oct 16 '17 at 17:20
  • @induction: You're welcome! :-) – Markus Scheuer Oct 16 '17 at 17:40