Closed form of partial hypergeometric sum

Question

Can we get closed form for $$\sum_{k=0}^m \left(-\frac12\right)^k \binom{2m}{m-k}k^p,\quad p\in\mathbb{N}\,?$$ In Concrete Mathematics Knuth describes Gosper's algorithm and its Zeilberger's extension, but they both fails on this partial hypergeometric sum. $-1/2$ can be replaced to $x$, but it doesn't help (I tried to differentiate it for any advances).

Hypergeometric function is not a closed form, however. Because this sum is a hypergeometric function yet (multiplied by binomial coeff.). It will be very cool to find closed form for $p=0$ or $p=1$. But if I understood correctly Knuth, there is no closed form.

Any ideas?

Answer 1

Let us write $$S_{m,p}(x)=\sum_{k=0}^m\binom{2m}{m-k}x^kk^p$$ You may try the generating function (ordinary for $m$, exponential for $p$) $$g(x,y,z)=\sum_{m,p}S_{m,p}(x)y^m\frac{z^p}{p!}.$$

First, for $p=0$ we have $$g(x,y,0)=\sum_m S_{m,0}(x)y^m=\sum_{0\le k\le m} \binom{2m}{m-k}x^ky^m$$ We can use $$\sum_m\binom{2m}{m-k}y^k=\frac1{\sqrt{1-4y}}\left(\frac{1-\sqrt{1-4y}}{1+\sqrt{1-4y}}\right)^k=\frac{1}{\sqrt{1-4y}}\frac{(1-\sqrt{1-4y})^{2k}}{4^ky^k}$$ and we get $$g(x,y,0)=\sum_mS_{m,0}(x)y^m=\frac1{\sqrt{1-4y}}\frac{4y}{4y-x(-1+\sqrt{1-4y})^2}.$$

For arbitrary $p$ $$g(x,y,z)=\sum_{m,p}\sum_{k=0}^mx^kk^p\binom{2m}{m-k}\,y^m\frac{z^p}{p!}=\sum_m\sum_{k=0}^m x^k \mathrm e^{kz}\binom{2m}{m-k}y^m=g(x\mathrm e^{z},y,0).$$

There are no hypergeometric functions. The result for $m$ and $p$ fixed is the coefficient of $y^mz^p$ multiplied by $p!$ in $g(x,y,z)$, it is written $$p!\left[y^mz^p\right]g(x,y,z).$$

Probably not what you hoped, but it is not likely to be improved into a less hypergeometric form.

Answer 2

One method that canbe used is the following.

Starting with $\delta = x D$ where $D = \frac{d}{dx}$ then $\delta^{p} x^{k} = k^{p} x^{k}$ for which \begin{align} S_{m} = \sum_{k=0}^{m} \binom{2m}{m-k} \, k^{p} \, x^{k} = \delta^{p} \, \sum_{k=0}^{m} \binom{2m}{m-k} \, x^{k} = \delta^{p} \, \sum_{k=0}^{m} \binom{2m}{k} \, x^{m-k}. \end{align} Since \begin{align} \sum_{k=0}^{m} \binom{2m}{k} \, y^{k} = (1+y)^{2m} - \binom{2m}{m+1} \, y^{m+1} \, {}_{2}F_{1}(1, 1-m; m+2; - y) \end{align} then \begin{align} S_{m} = \delta^{p} \left[ x^{m} \left(1 + \frac{1}{x} \right)^{2m} - \binom{2m}{m+1} \frac{1}{x} \, {}_{2}F{1}\left(1, 1-m; m+2; - \frac{1}{x} \right) \right]. \end{align} Calculating the derivatives may lead to what is desired.