<s>[INST] 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? 
 [/INST] 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
$$
\binom{m-1+\alpha}{m}+\binom{m-1+\alpha}{m-1}=\binom{m+\alpha}{m}\tag4
$$
</s>