Sum over k of r-kt Choose k by r over r-kt by s-(n-k)t Choose n-k by s over s-(n-k)t

Theorem
For $n \in \Z_{\ge 0}$:


 * $\displaystyle \sum_k A_k \left({r, t}\right) A_{n - k} \left({s, t}\right) = A_n \left({r + s, t}\right)$

where $A_n \left({x, t}\right)$ is the polynomial of degree $n$ defined as:
 * $A_n \left({x, t}\right) = \dbinom {x - n t} n \dfrac x {x - n t}$

where $x \ne n t$.

Proof
Let:
 * $\displaystyle S = \sum_k A_k \left({r, t}\right) A_{n - k} \left({s, t}\right)$

Both sides of the statement of the theorem are polynomials in $r$, $s$ and $t$.

Therefore it can be assumed that $r \ne k t \ne s$ for $0 \le k \le n$ or something will become undefined.

By replacing the polynomials $A_n$ with their binomial coefficient definitions, the theorem can be expressed as:
 * $\displaystyle S = \sum_k \dbinom {r - k t} k \dbinom {s - \left({n - k}\right) t} {n - k} \dfrac r {r - k t} \dfrac s {s - \left({n - k}\right) t}$

Using the technique of partial fractions:


 * $\dfrac 1 {r - k t} \dfrac 1 {s - \left({n - k}\right) t} = \dfrac 1 {r + s - n t} \left({\dfrac 1 {r - k t} + \dfrac 1 {s - \left({n - k}\right) t} }\right)$

Thus:
 * $\displaystyle S = \frac s {r + s - n t} \sum_k \dbinom {r - k t} k \dbinom {s - \left({n - k}\right) t} {n - k} \dfrac r {r - k t} + \frac r {r + s - n t} \sum_k \dbinom {r - k t} k \dbinom {s - \left({n - k}\right) t} {n - k} \dfrac s {s - \left({n - k}\right) t}$

From Sum over $k$ of $\dbinom {r - t k} k \dbinom {s - t \left({n - k}\right)} {n - k} \dfrac r {r - t k}$:
 * $\displaystyle \sum_{k \mathop \ge 0} \binom {r - t k} k \binom {s - t \left({n - k}\right)} {n - k} \frac r {r - t k} = \binom {r + s - t n} n$

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

Thus we have:
 * $S = \dfrac s {r + s - n t} \dbinom {r + s - t n} n + \dfrac r {r + s - n t} \dbinom {s + r - t n} n$

after changing $k$ to $n - k$ in the second term.

That is:
 * $S = \dbinom {r + s - n t} n \dfrac {r + s} {r + s - n t}$

which is $A_n \left({r + s, t}\right)$.