Half of the binomial theorem 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.
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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$\ds{\mrm{B}\pars{x} \equiv \sum_{k = 0}^{n}
{2n + 1 \choose k}x^{k}}$.

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}

Then,

\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}

Therefore,

\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}

The last integral is related to the
  Euler Type Hypergeometric Function. Namely,

$$
\int_{0}^{1}t^{n}\pars{1 + xt}^{-2n - 2}\,\dd t =
{1 \over n + 1}
\,\mbox{}_{2}\mrm{F}_{1}\pars{2n + 2,n +1;n + 2;-x}
$$

Finally

\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}
