# Sum over k to n of Stirling Number of the Second Kind of k with m by m+1^n-k

## Theorem

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

$\displaystyle \sum_{k \mathop \le n} \left\{ {k \atop m}\right\} \left({m + 1}\right)^{n - k} = \left\{ {n + 1 \atop m + 1}\right\}$

where $\displaystyle \left\{ {k \atop m}\right\}$ etc. denotes a Stirling number of the second kind.

## Proof

The proof proceeds by induction on $m$.

For all $m \in \Z_{\ge 0}$, let $P \left({m}\right)$ be the proposition:

$\displaystyle \forall n \in \Z_{\ge 0}: \sum_{k \mathop \le n} \left\{ {k \atop m}\right\} \left({m + 1}\right)^{n - k} = \left\{ {n + 1 \atop m + 1}\right\}$

### Basis for the Induction

$P \left({0}\right)$ is the case:

 $\displaystyle$  $\displaystyle \sum_{k \mathop \le n} \left\{ {k \atop 0}\right\} \left({0 + 1}\right)^{n - k}$ $\displaystyle$ $=$ $\displaystyle \sum_{k \mathop \le n} \delta_{k 0} \left({1}\right)^{n - k}$ Definition of Stirling Numbers of the Second Kind $\displaystyle$ $=$ $\displaystyle 1$ All terms but where $k = 0$ vanish $\displaystyle$ $=$ $\displaystyle \left\{ {n + 1 \atop m + 1}\right\}$ Stirling Number of the Second Kind of n+1 with 1

So $P \left({0}\right)$ is seen to hold.

This is the basis for the induction.

### Induction Hypothesis

Now it needs to be shown that, if $P \left({r}\right)$ is true, where $r \ge 1$, then it logically follows that $P \left({r + 1}\right)$ is true.

So this is the induction hypothesis:

$\displaystyle \sum_{k \mathop \le n} \left\{ {k \atop r}\right\} \left({r + 1}\right)^{n - k} = \left\{ {n + 1 \atop r + 1}\right\}$

from which it is to be shown that:

$\displaystyle \sum_{k \mathop \le n} \left\{ {k \atop r + 1}\right\} \left({r + 2}\right)^{n - k} = \left\{ {n + 1 \atop r + 2}\right\}$

### Induction Step

This is the induction step:

 $\displaystyle$  $\displaystyle \sum_{k \mathop \le n} \left\{ {k \atop r + 1}\right\} \left({r + 2}\right)^{n - k}$ $\displaystyle$ $=$ $\displaystyle \sum_{k \mathop \le n} \left({\left({r + 1}\right) \left\{ {k - 1 \atop r + 1}\right\} + \left\{ {k - 1 \atop r}\right\} }\right) \left({r + 2}\right)^{n - k}$ Definition of Stirling Numbers of the Second Kind $\displaystyle$ $=$ $\displaystyle \left({r + 1}\right) \sum_{k \mathop \le n} \left\{ {k - 1 \atop r + 1}\right\} \left({r + 2}\right)^{n - k} + \sum_{k \mathop \le n} \left\{ {k - 1 \atop r}\right\} \left({r + 2}\right)^{n - k}$

So $P \left({r}\right) \implies P \left({r + 1}\right)$ and the result follows by the Principle of Mathematical Induction.

Therefore:

$\displaystyle \forall m, n \in \Z_{\ge 0}: \sum_{k \mathop \le n} \left\{ {k \atop m}\right\} \left({m + 1}\right)^{n - k} = \left\{ {n + 1 \atop m + 1}\right\}$

$\blacksquare$