Chu-Vandermonde Identity/Proof 4

Theorem

\(\ds \sum_{k \mathop = 0}^n \binom r k \binom s {n - k}\) \(=\) \(\ds \binom r 0 \binom s n + \binom r 1 \binom s {n - 1} + \binom r 2 \binom s {n - 2} + \cdots + \binom r n \binom s 0\)
\(\ds \) \(=\) \(\ds \binom {r + s} n\)


Proof

From Sum over $k$ of $\dbinom {r - t k} k \dbinom {s - t \paren{n - k}} {n - k} \dfrac r {r - t k}$:


Let $r, s, t \in \R, n \in \Z$.

Then:

$\ds \sum_{k \mathop \ge 0} \binom {r - t k} k \binom {s - t \paren {n - k} } {n - k} \frac r {r - t k} = \binom {r + s - t n} n$


where $r, s, t \in \R, n \in \Z$.


Setting $t = 0$:

$\ds \sum_{k \mathop \ge 0} \binom r k \binom s {n - k} = \binom {r + s} n$

which is the result required.

$\blacksquare$


Source of Name

This entry was named for Chu Shih-chieh and Alexandre-Théophile Vandermonde.


Sources

  • 1997: Donald E. Knuth: The Art of Computer Programming: Volume 1: Fundamental Algorithms (3rd ed.) ... (previous) ... (next): $\S 1.2.6$: Binomial Coefficients: $\text {I}$
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