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I'm trying to find a general closed-form expression for the following sum:

$$m_p(N) = \sum_{n=0}^{N} \binom{N}{n} n^p$$

where $N,p\ge 0$ are integers.

So far I've been able to evaluate:

$$ \begin{eqnarray*} \sum_n \binom{N}{n} n^0 & = & 2^N\\ \sum_n \binom{N}{n} n^1 & = & 2^{N - 1} N\\ \sum_n \binom{N}{n} n^2 & = & 2^{N - 2} (N + 1) N\\ \sum_n \binom{N}{n} n^3 & = & 2^{N - 3} (N + 3) N^2 \end{eqnarray*} $$

but I don't see a general pattern here.

Is there a general analytical expression?

Note that this is not a textbook exercise. I'm not sure if a solution exists.

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  • $\begingroup$ There is a formula which uses the Stirling numbers of both kinds, or Stirling numbers of the second kind and falling factorials. Use the formula $n^p = \sum {p\brace k}(n)_k$, then use the technique in drhab’s answer. If you like, you can use $(N)_k=\sum {k\brace i}N^i$ to get a formula in powers of $N$. $\endgroup$ – Mike Earnest Jan 17 at 17:00
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The following equality might be helpful:

$$\sum_n\binom{N}{n}\left(n\right)_{p}=2^{N-p}\left(N\right)_{p}$$

Here $\left(x\right)_{p}$ is a notation for $x\left(x-1\right)\cdots\left(x-p+1\right)$.

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