How to prove this combinatorial identity? I am wondering how to prove the following identity:
$$\sum_{i=0}^{n-r} \frac{2^i (r+i) \binom{n-r}{i}}{(i+1) \binom{2n-r}{i+1}}=1?$$
It seems this might be related to the hypergeometric distribution, but I could not convert that form back into hypergeometric distribution form.
 A: Note that
$$
\begin{align}
\frac{2^i(r+i)\binom{n-r}{i}}{(i+1) \binom{2n-r}{i+1}}
&=\frac{2^i(r+i)}{n}\frac{\binom{2n-r-i-1}{n-1}}{\binom{2n-r}{n}}\tag{1}\\
&=\frac{2^i}{n}\frac{2n\binom{2n-r-i-1}{n-1}-n\binom{2n-r-i}{n}}{\binom{2n-r}{n}}\tag{2}\\
&=\frac{2^{i+1}\binom{2n-r-i-1}{n-1}-2^i\binom{2n-r-i}{n}}{\binom{2n-r}{n}}\tag{3}
\end{align}
$$
$(1)\quad$ $\frac1{i+1}\frac{\color{#C00000}{(n-r)!}}{i!\color{#C00000}{(n-r-i)!}}\frac{(i+1)!\color{#00A000}{(2n-r-i-1)!}}{\color{#00A000}{(2n-r)!}}=\frac1n\frac{n!\color{#C00000}{(n-r)!}}{\color{#00A000}{(2n-r)!}}\frac{\color{#00A000}{(2n-r-i-1)!}}{(n-1)!\color{#C00000}{(n-r-i)!}}$
$(2)\quad$ $\begin{array}{l}(2n-r-i)\binom{2n-r-i-1}{n-1}=n\binom{2n-r-i}{n}\\
\Rightarrow(r+i)\binom{2n-r-i-1}{n-1}=2n\binom{2n-r-i-1}{n-1}-n\binom{2n-r-i}{n}\end{array}$
$(3)\quad$ distribute $\frac{2^i}{n}$
Next, we have
$$
\begin{align}
\sum_{k=0}^{n-m}\binom{n-k}{m}2^k
&=\sum_{k=0}^{n-m}\sum_{j=0}^k\binom{n-k}{m}\binom{k}{j}\\
&=\sum_{j=0}^{n-m}\binom{n+1}{m+j+1}\tag{4}
\end{align}
$$
Applying $(4)$ to the sum of $(3)$, we get to a nicely telescoping sum:
$$
\begin{align}
{\large\sum_{i=0}^{n-r}}\;\frac{2^i(r+i)\binom{n-r}{i}}{(i+1) \binom{2n-r}{i+1}}
&={\large\sum_{i=0}^{n-r}}\;\frac{2^{i+1}\binom{2n-r-i-1}{n-1}-2^i\binom{2n-r-i}{n}}{\binom{2n-r}{n}}\\
&=\frac1{\binom{2n-r}{n}}\sum_{i=0}^{n-r}\left(2\binom{2n-r}{n+i}-\binom{2n-r+1}{n+i+1}\right)\\
&=\frac1{\binom{2n-r}{n}}\sum_{i=0}^{n-r}\left(\binom{2n-r}{n+i}-\binom{2n-r}{n+i+1}\right)\\
&=\frac1{\binom{2n-r}{n}}\binom{2n-r}{n}\\[6pt]
&=1\tag{5}
\end{align}
$$
