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 $m, n \in \Z_{\ge 0}$.

$\displaystyle \sum_{k \mathop \le n} {k \brace m} \paren {m + 1}^{n - k} = {{n + 1} \brace {m + 1}}$

where $\displaystyle {k \brace m}$ 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:

$\displaystyle \forall n \in \Z_{\ge 0}: \displaystyle \sum_{k \mathop \le n} {k \brace m} \paren {m + 1}^{n - k} = {{n + 1} \brace {m + 1}}$


Basis for the Induction

$\map P 0$ is the case:

\(\displaystyle \) \(\) \(\displaystyle \sum_{k \mathop \le 0} {k \brace m} \paren {m + 1}^{0 - k}\)
\(\displaystyle \) \(=\) \(\displaystyle {0 \brace m} \paren {m + 1}^{0 - 0}\)
\(\displaystyle \) \(=\) \(\displaystyle \delta_{0 m}\) Definition of Stirling Numbers of the Second Kind
\(\displaystyle \) \(=\) \(\displaystyle \delta_{1 \paren {m + 1} }\)
\(\displaystyle \) \(=\) \(\displaystyle {1 \brace {m + 1} }\) 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:

$\displaystyle \sum_{k \mathop \le r} {k \brace m} \paren {m + 1}^{r - k} = {{r + 1} \brace {m + 1}}$


from which it is to be shown that:

$\displaystyle \sum_{k \mathop \le {r + 1} } {k \brace m} \paren {m + 1}^{r + 1 - k} = { {r + 2} \brace {m + 1} }$


Induction Step

This is the induction step:

\(\displaystyle \) \(\) \(\displaystyle \sum_{k \mathop \le {r + 1} } {k \brace m} \paren {m + 1}^{r + 1 - k}\)
\(\displaystyle \) \(=\) \(\displaystyle { {r + 1} \brace m} \paren {m + 1}^{r + 1 - r - 1} + \paren {m + 1} \paren {\sum_{k \mathop \le r} {k \brace m} \paren {m + 1}^{r - k} }\)
\(\displaystyle \) \(=\) \(\displaystyle { {r + 1} \brace m} + \paren {m + 1} { {r + 1} \brace {m + 1} }\)
\(\displaystyle \) \(=\) \(\displaystyle { {r + 2} \brace {m + 1} }\) 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:

$\displaystyle \sum_{k \mathop \le n} {k \brace m} \paren {m + 1}^{n - k} = {{n + 1} \brace {m + 1}}$

$\blacksquare$


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