For all $n \in \Z$, let $\map P n$ be the proposition: :$\displaystyle \sum_{k \mathop \ge 0} \binom {r - t k} k \binom {s - t \paren {n - k} } {n - k} \frac r {r - t k} = \binom {r + s - t n} n$ Let the {{LHS}} of this equation be denoted $\tuple {r, s, t, n}$. Let $n = 0$. Then: {{begin-eqn}} {{eqn | o =        | r = \tuple {r, s, t, 0} }} {{eqn | r = \sum_{k \mathop \ge 0} \binom {r - t k} k \binom {s + t k} {- k} \frac r {r - t k}       | c =  }} {{eqn | r = \sum_{k \mathop \ge 0} \binom {r - t k} k \binom {s + t k} {- k} \frac r {r - t k} \delta_{k 0}       | c = N Choose Negative Number is Zero: $\dbinom {s + t k} {- k} = 0$ for all $k > 0$ }} {{eqn | r = \binom r 0 \binom s 0       | c = {{Defof|Kronecker Delta}} }} {{eqn | r = 1       | c = Binomial Coefficient with Zero }} {{eqn | r = \binom {r + s} 0       | c = Binomial Coefficient with Zero }} {{eqn | r = \binom {r + s - t n} n       | c = putting $n = 0$ }} {{end-eqn}} Thus $\map P 0$ holds. Let $n < 0$. Let $n = -m$ where $m > 0$. Then: {{begin-eqn}} {{eqn | o =        | r = \tuple {r, s, t, n} }} {{eqn | r = \sum_{k \mathop \ge 0} \binom {r - t k} k \binom {s - t \paren {-m - k} } {-m - k} \frac r {r - t k}       | c =  }} {{eqn | r = 0       | c = N Choose Negative Number is Zero: $\dbinom {s - t \paren {-m - k} } {-m - k} = 0$ }} {{eqn | r = \binom {r + s + t m} {-m}       | c = N Choose Negative Number is Zero: $\dbinom {r + s + t m} {-m} = 0$ for all $m$ }} {{end-eqn}} Thus $\map P n$ holds for all $n < 0$. It remains to demonstrate that $\map P n$ holds for all $n > 0$. The proof continues by strong induction on $n$. The procedure is to substitute $n - r + n t + m$ for the variable $s$ and establish that the identity holds for all $m \ge 0$. For all $m \in \Z_{\ge 0}$, let $\map P m$ be the proposition: :$\displaystyle \sum_{k \mathop \ge 0} \binom {r - t k} k \binom {\paren {n - r + n t + m} - t \paren {n - k} } {n - k} \frac r {r - t k} = \binom {r + \paren {n - r + n t + m} - t n} n$ That is: :$\displaystyle \sum_{k \mathop \ge 0} \binom {r - t k} k \binom {n - r + m + t k} {n - k} \frac r {r - t k} = \binom {n + m} n$ === Basis for the Induction === Consider the special case where $s = n - 1 - r + n t$. {{:Sum over k of r-tk Choose k by s-t(n-k) Choose n-k by r over r-tk/Proof 1/Basis for the Induction}}{{qed|lemma}} This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that, if $\map P j$ is true, for all $j$ such that $0 \le j \le m$, then it logically follows that $\map P {m + 1}$ is true. This is the induction hypothesis: :$\displaystyle \sum_{k \mathop \ge 0} \binom {r - t k} k \binom {n - r + m + t k} {n - k} \frac r {r - t k} = \binom {n + m} n$ from which it is to be shown that: :$\displaystyle \sum_{k \mathop \ge 0} \binom {r - t k} k \binom {n - r + m + 1 + t k} {n - k} \frac r {r - t k} = \binom {n + m + 1} n$ === Lemma === First a lemma: {{:Sum over k of r-tk Choose k by s-t(n-k) Choose n-k by r over r-tk/Proof 1/Lemma}} === Induction Step === This is the induction step: We have that $\tuple {r, n - 1 - r + n t, t, n}$ holds. We have also determined that if: :$\tuple {r, s, t, n}$ holds and: :$\tuple {r, s - t, t, n - 1}$ holds then: :$\tuple {r, s + 1, t, n}$ holds. So $\map P m \implies \map P {m + 1}$. Thus $\tuple {r, s, t, n}$ is shown to hold for infinitely many $s$. As both the {{LHS}} and {{RHS}} are polynomials in $s$ it follows that $\tuple {r, s, t, n}$ holds for all $s$. Therefore: :$\displaystyle \sum_{k \mathop \ge 0} \binom {r - t k} k \binom {s - t \left({n - k}\right)} {n - k} \frac r {r - t k} = \binom {r + s - t n} n$ for all $r, s, t \in \R, n \in \Z$. {{qed}}