Sum over k to n of Stirling Number of the Second Kind of k with m by m+1^n-k
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Theorem
Let $m, n \in \Z_{\ge 0}$.
- $\ds \sum_{k \mathop \le n} {k \brace m} \paren {m + 1}^{n - k} = {n + 1 \brace m + 1}$
where $\ds {k \brace m}$ etc. denotes a Stirling number of the second kind.
Proof
The proof proceeds by induction on $n$.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
- $\ds \forall n \in \Z_{\ge 0}: \sum_{k \mathop \le n} {k \brace m} \paren {m + 1}^{n - k} = {n + 1 \brace m + 1}$
Basis for the Induction
$\map P 0$ is the case:
\(\ds \sum_{k \mathop \le 0} {k \brace m} \paren {m + 1}^{0 - k}\) | \(=\) | \(\ds {0 \brace m} \paren {m + 1}^{0 - 0}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \delta_{0 m}\) | Definition of Stirling Numbers of the Second Kind | |||||||||||
\(\ds \) | \(=\) | \(\ds \delta_{1 \paren m + 1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds {1 \brace m + 1}\) | Stirling Number of the Second Kind of 1 |
So $\map P 0$ is seen to hold.
This is the basis for the induction.
Induction Hypothesis
Now it needs to be shown that, if $\map P r$ is true, where $r \ge 1$, then it logically follows that $\map P {r + 1}$ is true.
So this is the induction hypothesis:
- $\ds \sum_{k \mathop \le r} {k \brace m} \paren {m + 1}^{r - k} = {r + 1 \brace m + 1}$
from which it is to be shown that:
- $\ds \sum_{k \mathop \le {r + 1} } {k \brace m} \paren {m + 1}^{r + 1 - k} = {r + 2 \brace m + 1}$
Induction Step
This is the induction step:
\(\ds \sum_{k \mathop \le {r + 1} } {k \brace m} \paren {m + 1}^{r + 1 - k}\) | \(=\) | \(\ds {r + 1 \brace m} \paren {m + 1}^{r + 1 - r - 1} + \paren {m + 1} \paren {\sum_{k \mathop \le r} {k \brace m} \paren {m + 1}^{r - k} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds {r + 1 \brace m} + \paren {m + 1} {r + 1 \brace m + 1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds {r + 2 \brace m + 1}\) | Definition of Stirling Numbers of the Second Kind |
So $\map P r \implies \map P {r + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
- $\ds \sum_{k \mathop \le n} {k \brace m} \paren {m + 1}^{n - k} = {n + 1 \brace m + 1}$
$\blacksquare$
Sources
- 1997: Donald E. Knuth: The Art of Computer Programming: Volume 1: Fundamental Algorithms (3rd ed.) ... (previous) ... (next): $\S 1.2.6$: Binomial Coefficients: $(56)$