Sum over k of r Choose k by s+k Choose n by -1^r-k
Theorem
Let $s \in \R, r \in \Z_{\ge 0}, n \in \Z$.
Then:
- $\ds \sum_k \binom r k \binom {s + k} n \paren {-1}^{r - k} = \binom s {n - r}$
where $\dbinom r k$ etc. are binomial coefficients.
Corollary
- $\ds \sum_k \binom r k \binom k n \paren {-1}^{r - k} = \delta_{n r}$
where $\delta_{n r}$ is the Kronecker delta.
Proof
The proof proceeds by induction on $r$.
For all $r \in \Z_{>0}$, let $\map P r$ be the proposition:
- $\ds \sum_k \binom r k \binom {s + k} n \paren {-1}^{r - k} = \binom s {n - r}$
Basis for the Induction
$\map P 0$ is the case:
\(\ds \sum_k \binom 0 k \binom {s + k} n \paren {-1}^{0 - k}\) | \(=\) | \(\ds \delta_{0 k} \binom {s + k} n \paren {-1}^{0 - k}\) | Zero Choose n | |||||||||||
\(\ds \) | \(=\) | \(\ds \binom s n\) | All terms vanish but for $k = 0$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \binom s {n - 0}\) |
Thus $\map P 0$ is seen to hold.
This is the basis for the induction.
Induction Hypothesis
Now it needs to be shown that, if $\map P m$ is true, where $m \ge 0$, then it logically follows that $\map P {m + 1}$ is true.
This is the induction hypothesis:
- $\ds \sum_k \binom m k \binom {s + k} n \paren {-1}^{m - k} = \binom s {n - m}$
from which it is to be shown that:
- $\ds \sum_k \binom {m + 1} k \binom {s + k} n \paren {-1}^{m + 1 - k} = \binom s {n - \paren {m + 1} }$
Induction Step
This is the induction step:
\(\ds \) | \(\) | \(\ds \sum_k \binom {m + 1} k \binom {s + k} n \paren {-1}^{m + 1 - k}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sum_k \paren {\binom m k + \binom m {k - 1} } \binom {s + k} n \paren {-1}^{m + 1 - k}\) | Pascal's Rule | |||||||||||
\(\ds \) | \(=\) | \(\ds \sum_k \binom m k \binom {s + k} n \paren {-1}^{m + 1 - k} + \sum_k \binom m {k - 1} \binom {s + k} n \paren {-1}^{m + 1 - k}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds -\sum_k \binom m k \binom {s + k} n \paren {-1}^{m - k} + \sum_k \binom m {k - 1} \binom {s + k} n \paren {-1}^{m + 1 - k}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds -\binom s {n - m} + \sum_k \binom m {k - 1} \binom {s + k} n \paren {-1}^{m + 1 - k}\) | Induction Hypothesis | |||||||||||
\(\ds \) | \(=\) | \(\ds -\binom s {n - m} + \sum_k \binom m {k - 1} \paren {\binom {s + k - 1} n + \binom {s + k - 1} {n - 1} } \paren {-1}^{m + 1 - k}\) | Pascal's Rule | |||||||||||
\(\ds \) | \(=\) | \(\ds -\binom s {n - m} + \sum_k \binom m {k - 1} \binom {s + k - 1} n \paren {-1}^{m - \paren {k - 1} }\) | ||||||||||||
\(\ds \) | \(\) | \(\, \ds + \, \) | \(\ds \sum_k \binom m {k - 1} \binom {s + k - 1} {n - 1} \paren {-1}^{m - \paren {k - 1} }\) | |||||||||||
\(\ds \) | \(=\) | \(\ds -\binom s {n - m} + \sum_k \binom m k \binom {s + k} n \paren {-1}^{m - k} + \sum_k \binom m k \binom {s + k} {n - 1} \paren {-1}^{m - k}\) | Translation of Index Variable of Summation | |||||||||||
\(\ds \) | \(=\) | \(\ds -\binom s {n - m} + \binom s {n - m} + \binom s {n - 1 - m}\) | Induction Hypothesis | |||||||||||
\(\ds \) | \(=\) | \(\ds \binom s {n - \paren {m + 1} }\) | simplifying |
So $\map P m \implies \map P {m + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
- $\ds \sum_k \binom r k \binom {s + k} n \paren {-1}^{r - k} = \binom s {n - r}$
for all $s \in \R, r \in \Z_{\ge 0}, n \in \Z$.
$\blacksquare$
Sources
- 1997: Donald E. Knuth: The Art of Computer Programming: Volume 1: Fundamental Algorithms (3rd ed.) ... (previous) ... (next): $\S 1.2.6$: Binomial Coefficients: $\text {I} \ (23)$