Is there a closed-form expression for the sum $\sum_{k=0}^n\binom{n}kk^p$ given positive integers $n,\,p$? Earlier I thought of this series but failed to figure out a closed-form expression in $n,\,p$ (other than the trivial case $p=0$).
$$p=0\colon\,\sum_{k=0}^n\binom{n}kk^0=2^n$$
I know that $\sum_{k=0}^n\binom{n}k=2^n$ and $\sum_{k=0}^nk^n=\frac{k^{n+1}-1}{k-1}$ but I am unsure of whether these would be of much use now.
Additionally, what about the similar series $\sum_{k=0}^n\binom{n}kk^n$ where $p=n$?
Best Answer
If you consider $p$ as fixed, then the below can be considered as closed form I suppose:
$$\sum_{k=0}^{n} \binom{n}{k} k^p = \sum_{k=1}^{p} s(p,k) n(n-1)\dots(n-k+1) 2^{n-k} \quad \quad (1)$$
where $s(k,p)$ is a stirling number of the second kind.
If you denote the operator of differentiating and multiplying by $x$ as $D_{x}$
Then we have that
$$(D_{x})^{n}f(x) = \sum_{k=1}^{n} s(n,k) f^{k}(x) x^{k}$$
where $s(n,k)$ is the stirling number of the second kind and $f^k(x)$ is the $k^{th}$ derivative of $f(x)$.
This can easily be proven using the identity $$s(n,k) = s(n-1,k-1) + k \cdot s(n-1,k)$$
To prove $(1)$ above, we apply $D_x$, $p$ times to $(1+x)^n$, and set $x=1$.