Chu-Vandermonde Identity/Proof 4

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Theorem

$\displaystyle \sum_k \binom r k \binom s {n - k} = \binom {r + s} n$


Proof

From Sum over $k$ of $\dbinom {r - t k} k \dbinom {s - t \left({n - k}\right)} {n - k} \dfrac r {r - t k}$:


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

Then:

$\displaystyle \sum_{k \mathop \ge 0} \binom {r - t k} k \binom {s - t \left({n - k}\right)} {n - k} \frac r {r - t k} = \binom {r + s - t n} n$


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


Setting $t = 0$:

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

which is the result required.

$\blacksquare$


Source of Name

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


Sources