Sum over k of r+tk choose k by s-tk choose n-k
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Theorem
Let $n \in \Z_{\ge 0}$ be a non-negative integer.
Then:
- $\ds \sum_k \dbinom {r + t k} k \dbinom {s - t k} {n - k} = \sum_{k \mathop \ge 0} \dbinom {r + s - k} {n - k} t^k$
where $\dbinom {r + t k} k$ and so on denotes a binomial coefficient.
Proof
Let $\map f {r, s, t, n}$ be the function defined as:
- $\ds \map f {r, s, t, n} := \sum_k \dbinom {r + t k} k \dbinom {s - t k} {n - k}$
We have:
\(\ds \) | \(\) | \(\ds \sum_k \dbinom {r + t k} k \dbinom {s - t k} {n - k}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sum_k \dbinom {r + t k} k \dbinom {s - t k} {n - k} \frac {r + t k} {r + t k}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sum_k \dbinom {r + t k} k \dbinom {s - t k} {n - k} \frac r {r + t k} + \sum_k \dbinom {r + t k} k \dbinom {s - t k} {n - k} \frac {t k} {r + t k}\) |
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Sources
- 1997: Donald E. Knuth: The Art of Computer Programming: Volume 1: Fundamental Algorithms (3rd ed.) ... (previous) ... (next): $\S 1.2.6$: Binomial Coefficients: Exercise $28$