Sum over k of Stirling Numbers of the Second Kind of k with m by n choose k

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Theorem

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

$\ds \sum_k {k \brace m} \binom n k = {n + 1 \brace m + 1}$

where:

$\ds {k \brace m}$ denotes a Stirling number of the second kind
$\dbinom n k$ denotes a binomial coefficient.


Proof

The proof proceeds by induction.


For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:

$\ds \forall m \in \Z_{\ge 0}: \sum_k {k \brace m} \binom n k = {n + 1 \brace m + 1}$


Basis for the Induction

$\map P 0$ is the case:

\(\ds \sum_k {k \brace m} \binom 0 k\) \(=\) \(\ds \sum_k {k \brace m} \delta_{0 k}\) Zero Choose n
\(\ds \) \(=\) \(\ds {0 \brace m}\) all terms vanish except for $k = 0$
\(\ds \) \(=\) \(\ds \delta_{0 m}\) Stirling Number of the Second Kind of 0
\(\ds \) \(=\) \(\ds \delta_{1 \paren {m + 1} }\) Definition of Kronecker Delta
\(\ds \) \(=\) \(\ds {1 \brace m + 1}\) Stirling Number of the Second Kind of 1
\(\ds \) \(=\) \(\ds {0 + 1 \brace m + 1}\)

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:

$\ds \sum_k {k \brace m} \binom r k = {r + 1 \brace m + 1}$


from which it is to be shown that:

$\ds \sum_k {k \brace m} \binom {r + 1} k = {r + 2 \brace m + 1}$


Induction Step

This is the induction step:

\(\ds \) \(\) \(\ds \sum_k {k \brace m} \binom {r + 1} k\)
\(\ds \) \(=\) \(\ds \sum_k {k \brace m} \paren {\binom r k + \binom r {k - 1} }\) Pascal's Rule
\(\ds \) \(=\) \(\ds \sum_k {k \brace m} \binom r k + \sum_k {k \brace m} \binom r {k - 1}\)
\(\ds \) \(=\) \(\ds {r + 1 \brace m + 1} + \sum_k {k \brace m} \binom r {k - 1}\) Induction Hypothesis
\(\ds \) \(=\) \(\ds {r + 1 \brace m + 1} + \sum_k \paren {m {k - 1 \brace m} + {k - 1 \brace m - 1} } \binom r {k - 1}\) Definition of Stirling Numbers of the Second Kind
\(\ds \) \(=\) \(\ds {r + 1 \brace m + 1} + m \sum_k {k - 1 \brace m} \binom r {k - 1} + \sum_k {k - 1 \brace m - 1} \binom r {k - 1}\)
\(\ds \) \(=\) \(\ds {r + 1 \brace m + 1} + m \sum_k {k \brace m} \binom r k + \sum_k {k \brace m - 1} \binom r k\) Translation of Index Variable of Summation
\(\ds \) \(=\) \(\ds {r + 1 \brace m + 1} + m {r + 1 \brace m + 1} + {r + 1 \brace m}\) Induction Hypothesis
\(\ds \) \(=\) \(\ds \paren {m + 1} {r + 1 \brace m + 1} + {r + 1 \brace m}\)
\(\ds \) \(=\) \(\ds {r + 2 \brace m + 1}\) 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:

$\ds \forall m, n \in \Z_{\ge 0}: \sum_k {k \brace m} \binom n k = {n + 1 \brace m + 1}$

$\blacksquare$


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