The binomial theorem states that the generating function $\sum_{k=0}^n {n \choose k} x^k$ is equal to $(1+x)^n$ for any $n$. For a given $n$, let $$B(x)=\sum_{k=0}^n {2n+1\choose k} x^k.$$
That is, we sum over only the first half of the binomial coefficients. Does $B(x)$ have a nice closed expression?
Currently the only idea I have is the observation that $$(1+x)^{2n+1}=\sum_{k=0}^n {2n+1\choose k} x^k+\sum_{k=0}^n {2n+1 \choose k} x^{2n+1-k}=B(x)+x^{2n+1}B(1/x),$$
but from there I'm not sure how to proceed.
I should also note that the "real" problem I'm ultimately interested in is finding a formula for $$C(x)=\sum_{k=0}^{n-1} A(2n,k) x^k,$$ where $A(n,k)$ is the Eulerian number. I would be more than happy with just an answer to this problem, but it seems like any solution here should be at least as difficult as a solution for the binomial problem.
Best Answer
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
With $\ds{\mrm{B}\pars{0} = 1}$: \begin{align} \mrm{B}'\pars{x} & \equiv \sum_{k = 0}^{n}{2n + 1\choose k}kx^{k - 1} = \sum_{k = 1}^{n}{\pars{2n + 1}! \over \pars{k - 1}!\pars{2n + 1 - k}!}\,x^{k - 1} \\[5mm] & = \sum_{k = 0}^{n - 1}{\pars{2n + 1}! \over k!\pars{2n - k}}\,x^{k} = \sum_{k = 0}^{n - 1}\pars{2n + 1 - k} {2n + 1 \choose k}x^{k} \\[5mm] & = -\pars{n + 1}{2n + 1 \choose n}x^{n} + \sum_{k = 0}^{n}\pars{2n + 1 - k}{2n + 1 \choose k}x^{k} \\[5mm] & = -\pars{n + 1}{2n + 1 \choose n}x^{n} + \pars{2n + 1}\mrm{B}\pars{x} - x\,\mrm{B}'\pars{x} \end{align}
\begin{align} &\mrm{B}'\pars{x} - {2n + 1 \over 1 + x}\,\mrm{B}\pars{x} = -\pars{n + 1}{2n + 1 \choose n}{x^{n} \over 1 + x} \\[5mm] & \totald{\bracks{\pars{1 + x}^{-2n - 1}\,\mrm{B}\pars{x}}}{x} = -\pars{n + 1}{2n + 1 \choose n}{x^{n} \over \pars{1 + x}^{2n + 2}} \\[5mm] & {\mrm{B}\pars{x} \over \pars{1 + x}^{2n + 1}} - 1 = -\pars{n + 1}{2n + 1 \choose n} \int_{0}^{x}{t^{n} \over \pars{1 + t}^{2n + 2}}\,\dd t \end{align}
\begin{align} \mrm{B}\pars{x} & = \pars{1 + x}^{2n + 1} \\[2mm] & - \pars{n + 1}{2n + 1 \choose n}\pars{1 + x}^{2n + 1}x^{n + 1} \int_{0}^{1}t^{n}\pars{1 + xt}^{-2n - 2}\,\dd t \end{align}
$$ \int_{0}^{1}t^{n}\pars{1 + xt}^{-2n - 2}\,\dd t = {1 \over n + 1} \,\mbox{}_{2}\mrm{F}_{1}\pars{2n + 2,n +1;n + 2;-x} $$
\begin{align} \mrm{B}\pars{x} & \equiv \sum_{k = 0}^{n}{2n + 1\choose k}x^{k} \\[5mm] & = \pars{1 + x}^{2n + 1} \\[2mm] & - {2n + 1 \choose n}\pars{1 + x}^{2n + 1}x^{n + 1}\, \mbox{}_{2}\mrm{F}_{1}\pars{2n + 2,n +1;n + 2;-x} \end{align}