Summation of Central Binomial Coefficients divided by even powers of $2$ Whilst working out this problem the following summation emerged:
$$\sum_{m=0}^n\frac 1{2^{2m}}\binom {2m}m$$
The is equivalent to 
$$\begin{align}
\sum_{m=0}^n \frac {(2m-1)!!}{2m!!}&=\frac 12+\frac {1\cdot3}{2\cdot 4}+\frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6}+\cdots +\frac{1\cdot 3\cdot 5\cdot \cdots \cdot(2n-1)}{2\cdot 4\cdot 6\cdot \cdots \cdot 2n}\\
&=\frac 12\left(1+\frac 34\left(1+\frac 56\left(1+\cdots \left(1+\frac {2n-1}{2n}\right)\right)\right)\right)
\end{align}$$
 and terms are the same as coefficients in the expansion of $(1-x)^{-1/2}$.
Once the solution 
$$ \frac {n+1}{2^{2n+1}}\binom {2n+2}{n+1}$$
is known, the telescoping sum can be easily derived, i.e. 
  $$\frac 1{2^{2m}}\binom {2m}m=\frac {m+1}{2^{2(m+1)-1}}\binom {2(m+1)}{m+1}-\frac m{2^{2m-1}}\binom {2m}m$$
However, without knowing this a priori, how would we have approached this problem? 
 A: You may do the following to get rid of the powers of two and transform it into a standard sum:
\begin{equation} \binom{2m}{m}=\frac{(2m)!}{m!m!}=\frac{2^m m! 1\cdot3\cdot5\cdot \cdots (2m-1)}{m!m!}=2^{2m} \frac{(1/2)(3/2)\cdots (m-1/2)}{m!}=2^{2m}\binom{m-1/2}{m}~~~~~~ (1) \end{equation}
Then your sum becomes
$$\sum_{m=0}^{n} \binom{m-1/2}{m}=\binom{n-1/2 +1}{n} =\binom{n+1-1/2}{n}=\frac{n+1}{n+1-1/2-n}\frac{n+1-1/2-n}{n+1}\binom{n+1-1/2}{n}=\frac{n+1}{1/2}\binom{n+1-1/2}{n+1}=2(n+1)\frac{1}{2^{2(n+1)}}\binom{2(n+1)}{n+1}=\frac{n+1}{2^{2{n+1}}}\binom{2n+2}{n+1},$$
where the first equality is just the usual identity $\sum_{j=0}^n \binom{r+j}{j} =\binom{r+n+1}{n}$ valid for any real $r$ and any non-negative integer $n$, and the last few steps were just trying to slightly transform the coefficient in a way that allowed application of $(1)$.
A: Another approach. By Euler/De Moivre's formula we have
$$A_n=\frac{1}{4^n}\binom{2n}{n} = \frac{1}{2\pi}\int_{-\pi}^{\pi}\cos^{2n}(x)\,dx \tag{1}$$
since $\cos(x)^{2n} = \frac{1}{4^n}\sum_{j=0}^{2n}\binom{2n}{j}e^{(2n-j)ix}e^{-jix}$ and $\int_{-\pi}^{\pi}e^{kix}\,dx =2\pi \delta(k)$, hence only the contribute given by $j=n$ survives. It follows that
$$ \color{red}{\sum_{n=0}^{N}\frac{1}{4^n}\binom{2n}{n}}=\frac{1}{2\pi}\int_{-\pi}^{\pi}\frac{1-\cos^{2N+2}(x)}{\sin^2(x)}\,dx =(2N+2)\,A_{N+1}=\color{red}{\frac{N+1}{2^{2N+1}}\binom{2N+2}{N+1}}\tag{2}$$
by integration by parts ($\int\frac{dx}{\sin^2 x}=-\cot x$).
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,}
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 \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)}
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Indeed, we can use the result \begin{equation}
\left.\sum_{n = 0}^{\infty}{2n \choose n}x^{n}
\,\right\vert_{\ \verts{x}\ <\ 1/4} =
{1 \over \root{1 - 4x}}\label{1}\tag{1}
\end{equation}
to evaluate the finite sum whenever we don't know the ' telescopic stuff '. It requires the use of a generating function:

\begin{align}
\mc{F}\pars{z} & \equiv
\sum_{n = 0}^{\infty}z^{n}\bracks{%
\sum_{m = 0}^{n}{1 \over 2^{2m}}{2m \choose m}}
\\[5mm] &\iff
\sum_{m = 0}^{n}{1 \over 2^{2m}}{2m \choose m} = \bracks{z^{n}}\mc{F}\pars{z}\,,
\quad\color{#f00}{\verts{z} < 1}\label{2}\tag{2} 
\end{align}

\begin{align}
\mc{F}\pars{z} & \equiv
\sum_{n = 0}^{\infty}z^{n}\bracks{%
\sum_{m = 0}^{n}{1 \over 2^{2m}}{2m \choose m}} =
\sum_{m = 0}^{\infty}{1 \over 4^{m}}{2m \choose m}
\sum_{n = m}^{\infty}z^{n}
\\[5mm] & =
\bracks{\sum_{m = 0}^{\infty}\pars{z \over 4}^{m}{2m \choose m}}
\sum_{n = 0}^{\infty}z^{n}
\\[5mm] & =
{1 \over \root{1 - 4\pars{z/4}}}\,{1 \over 1 - z} = \pars{1 - z}^{-3/2}
\\[5mm] & =
\sum_{n = 0}^{\infty}{-3/2 \choose n}
\pars{-1}^{n}z^{n}
\qquad\pars{~\mbox{see expression}\ \eqref{1}~}
\\[5mm] & =
\sum_{n = 0}^{\infty}\bracks{{n + 1/2 \choose n}\pars{-1}^{n}}\pars{-1}^{n}z^{n} \end{align}

\begin{equation}
\mc{F}\pars{z} =
\sum_{n = 0}^{\infty}{n + 1/2 \choose n}z^{n}
\qquad\stackrel{\mrm{see\ expression}\ \eqref{2}}{\implies}\qquad
\bbox[15px,#ffe,border:1px dashed navy]{\ds{%
\sum_{m = 0}^{n}{1 \over 2^{2m}}{2m \choose m} = {n + 1/2 \choose n}}} \\
\end{equation}

The last expression can be rewritten as the OP reported answer
$\ds{{n + 1 \over 2^{2n + 1}}{2n + 2 \choose n + 1}}$ by means of the $\ds{\Gamma}$-Duplication Formula.
A: Using an Extension of Pascal's Rule
$$
\begin{align}
\sum_{m=0}^n\frac1{2^{2m}}\binom{2m}{m}
&=\sum_{m=0}^n\frac{(2m-1)!!}{(2m)!!}\tag{1a}\\
&=\sum_{m=0}^n\binom{m-\frac12}{m}\tag{1b}\\
&=\sum_{m=0}^n\left[\binom{m+\frac12}{m}-\binom{m-1+\frac12}{m-1}\right]\tag{1c}\\
&=\binom{n+\frac12}{n}\tag{1d}\\[6pt]
&=\frac{(2n+1)!!}{(2n)!!}\tag{1e}\\[6pt]
&=\frac{2n+1}{2^{2n}}\binom{2n}{n}\tag{1f}
\end{align}
$$
Explanation:
$\text{(1a)}$: $\frac{\color{#C00}{(2m)!}}{\color{#C00}{2^mm!}\,\color{#090}{2^mm!}}=\frac{\color{#C00}{(2m-1)!!}}{\color{#090}{(2m)!!}}$
$\text{(1b)}$: divide numerator and denominator by $2^m$
$\text{(1c)}$: apply $(4)$ with $\alpha=\frac12$
$\text{(1d)}$: telescoping sum
$\text{(1e)}$: multiply numerator and denominator by $2^n$
$\text{(1f)}$: $\frac{(2n+1)!!}{(2n)!!}=\frac{(2n+1)\color{#C00}{(2n-1)!!}}{\color{#090}{(2n)!!}}=\frac{(2n+1)\color{#C00}{(2n)!}}{\color{#C00}{2^nn!}\,\color{#090}{2^nn!}}$

Extension of Pascal's Rule
Newton's Generalized Binomial Theorem says
$$
\begin{align}
\sum_{m=0}^\infty\binom{m+\alpha}{m}x^m
&=\sum_{m=0}^\infty(-1)^m\binom{-1-\alpha}{m}x^m\tag{2a}\\
&=(1-x)^{-1-\alpha}\tag{2b}
\end{align}
$$
Explanation:
$\text{2a}$: convert to negative binomial coefficient
$\text{2b}$: Binomial Theorem
Thus,
$$
\begin{align}
\sum_{m=0}^\infty\binom{m-1+\alpha}{m}x^m
&=(1-x)^{-\alpha}\tag{3a}\\[6pt]
&=(1-x)(1-x)^{-1-\alpha}\tag{3b}\\[9pt]
&=(1-x)\sum_{m=0}^\infty\binom{m+\alpha}{m}x^m\tag{3c}\\
&=\sum_{m=0}^\infty\binom{m+\alpha}{m}\left(x^m-x^{m+1}\right)\tag{3d}\\[3pt]
&=\sum_{m=0}^\infty\left[\binom{m+\alpha}{m}-\binom{m-1+\alpha}{m-1}\right]x^m\tag{3e}
\end{align}
$$
Explanation:
$\text{(3a)}$: apply $\text{(2b)}$
$\text{(3b)}$: factor out $(1-x)$
$\text{(3c)}$: apply $\text{(2b)}$
$\text{(3d)}$: distribute $(1-x)$
$\text{(3e)}$: substitute $m\mapsto m-1$ in the subtrahend
Thus, for arbitrary $\alpha\in\mathbb{R}$, we can extend Pascal's Rule to
$$
\binom{m-1+\alpha}{m}+\binom{m-1+\alpha}{m-1}=\binom{m+\alpha}{m}\tag4
$$
