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.
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 $x\left(x-1\right)\cdots\left(x-p+1\right)$.