Binomial coefficients equality or maybe probability Let $m,n$ be positive integers. 
Evaluate the following expression:
$$
F(m,n) = \sum\limits_{i=0}^n\frac{\binom{m+i}{i}}{2^{m+i+1}}+
\sum\limits_{i=0}^m\frac{\binom{n+i}{i}}{2^{n+i+1}}.
$$
Calclulations give the hypothesis that $$F(m,n)=1,$$ for all positive integers $m,n$.
Also if $m=n$, then
$$
F(m,m) = \sum\limits_{i=0}^m\frac{\binom{m+i}{i}}{2^{m+i}} = \sum\limits_{i=0}^m\frac{\binom{m+i}{m}}{2^{m+i}}. 
$$
The numerator of every summand is equal to the number of $m$-subsets in $m+i$-set and denominator is equal to the number of subsets in $m+i$-set. So, I think it maybe the key to solution.
 A: Here is an answer in two steps based upon generating functions

First step: The following identity holds true for $m,n\geq 0$
  \begin{align*}
\sum_{i=0}^n\binom{m+i}{i}\frac{1}{2^{m+i+1}}=\frac{1}{2^{m+n+1}}\sum_{i=0}^n\binom{m+n+1}{i}\tag{1}
\end{align*}

In order to show (1) it is convenient to use the coefficient of operator $[z^i]$ to denote the coefficient of $z^i$. This way we can write e.g.
\begin{align*}
[z^i](1+z)^n=\binom{n}{i}
\end{align*}

We obtain
  \begin{align*}
\color{blue}{\sum_{i=0}^{n}\binom{m+i}{i}\frac{1}{2^{m+i+1}}}
&=\frac{1}{2^{m+n+1}}\sum_{i=0}^n\binom{m+n-i}{n-i}2^i\tag{2}\\
&=\frac{1}{2^{m+n+1}}\sum_{i=0}^\infty[z^{n-i}](1+z)^{m+n-i}2^i\tag{3}\\
&=\frac{1}{2^{m+n+1}}[z^n](1+z)^{m+n}\sum_{i=0}^\infty\left(\frac{2z}{1+z}\right)^i\tag{4}\\
&=\frac{1}{2^{m+n+1}}[z^n](1+z)^{m+n}\frac{1}{1-\frac{2z}{1+z}}\tag{5}\\
&=\frac{1}{2^{m+n+1}}[z^n]\frac{(1+z)^{m+n+1}}{1-z}\\
&=\frac{1}{2^{m+n+1}}\sum_{i=0}^n\binom{m+n+1}{i}[z^{n-i}]\frac{1}{1-z}\tag{6}\\
&\color{blue}{=\frac{1}{2^{m+n+1}}\sum_{i=0}^n\binom{m+n+1}{i}}\tag{7}
\end{align*}
  and (1) follows.

Comment:


*

*In (2) we exchange the order of summation by letting $i\rightarrow n-i$.

*In (3) we apply the coefficient of operator and set the upper limit of the sum to infty without changing anything since we are adding zeros only.

*In (4) we use the linearity of the coefficient of operator and apply the rule $$[z^{p-q}]A(z)=[z^p]z^qA(z)$$

*In (5) we use the geometric series expansion.

*In (6) we select the coefficient of $[z^{n-i}]$ in $(1+z)^{m+n+1}$ and restrict the upper limit of the sum to $n$ since the exponent of $z^{n-i}$ is non-negative.

*In (7) we note the coefficient of $[z^{n-i}]$ in $\frac{1}{1-z}=1+z+z^2+\cdots$ is $1$.

Second step: $F(m,n)$
Now it's time to harvest. We obtain from (7)
\begin{align*}
\color{blue}{\frac{1}{2^{m+n+1}}\sum_{i=0}^m\binom{m+n+1}{i}}
&=1-\frac{1}{2^{m+n+1}}\sum_{i=m+1}^{m+n+1}\binom{m+n+1}{i}\tag{8}\\
&=1-\frac{1}{2^{m+n+1}}\sum_{i=0}^{n}\binom{m+n+1}{n-i}\tag{9}\\
&\color{blue}{=1-\frac{1}{2^{m+n+1}}\sum_{i=0}^{n}\binom{m+n+1}{i}}\tag{10}\\
\end{align*}

Comment:


*

*In (8) we use $\sum_{i=0}^{p}\binom{p}{i}=2^p$.

*In (9) we shift the index to start with $i=0$ and we use the binomial identity $\binom{p}{q}=\binom{p}{p-q}$.

*In (10) we change the order of summation by letting $i\rightarrow n-i$.

We conclude from (7) and (10)
  \begin{align*}
\color{blue}{F(m,n)}&=\sum_{i=0}^{n}\binom{m+i}{i}\frac{1}{2^{m+i+1}}
+\sum_{i=0}^{m}\binom{n+i}{i}\frac{1}{2^{n+i+1}}\\
&=\frac{1}{2^{m+n+1}}\sum_{i=0}^n\binom{m+n+1}{i}
+\frac{1}{2^{m+n+1}}\sum_{i=0}^m\binom{m+n+1}{i}\\
&=\frac{1}{2^{m+n+1}}\sum_{i=0}^n\binom{m+n+1}{i}+\left(1-\frac{1}{2^{m+n+1}}\sum_{i=0}^{n}\binom{m+n+1}{i}\right)\\
&\color{blue}{=1}
\end{align*}

A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{\mrm{F}\pars{m,n} \equiv
\sum_{i = 0}^{n}{{m + i \choose i} \over 2^{m + i + 1}} +
\sum_{i = 0}^{m}{{n + i \choose i} \over 2^{n + i + 1}}.\qquad
\mrm{F}\pars{m,n} = 1:\ {\large ?}}$

\begin{equation}
\mbox{Note that}\quad\mrm{F}\pars{m,n} =
{1 \over 2^{m + 1}}\sum_{i = 0}^{n}{-m - 1 \choose i}\pars{-\,{1 \over 2}}^{i} +
{1 \over 2^{n + 1}}\sum_{i = 0}^{m}{-n - 1 \choose i}\pars{-\,{1 \over 2}}^{i}
\label{1}\tag{1}
\end{equation}

The Generation Function of the
$\ds{\color{#f00}{\texttt{first term}}}$, in \eqref{1}, is given by

$\ds{\pars{~\mbox{multiply by the factor}\ w^{m}z^{n}\ \mbox{and sum over}\
m,n \in \mathbb{N}_{\ \geq\ 0} ~}}$
\begin{align}
&\bbox[#ffe,15px]{\ds{\sum_{m = 0}^{\infty}\sum_{n = 0}^{\infty}w^{m}z^{n}
\bracks{{1 \over 2^{m + 1}}
\sum_{i = 0}^{n}{-m - 1 \choose i}\pars{-\,{1 \over 2}}^{i}}}}
 =
\sum_{m = 0}^{\infty}{w^{m} \over 2^{m + 1}}\sum_{i = 0}^{\infty}
{-m - 1 \choose i}\pars{-\,{1 \over 2}}^{i}
\sum_{n = i}^{\infty}z^{n}
\\[5mm] = &\
\sum_{m = 0}^{\infty}{w^{m} \over 2^{m + 1}}\sum_{i = 0}^{\infty}
{-m - 1 \choose i}\pars{-\,{1 \over 2}}^{i}
\sum_{n = 0}^{\infty}z^{n + i} =
{1 \over 1 - z}\sum_{m = 0}^{\infty}{w^{m} \over 2^{m + 1}}\sum_{i = 0}^{\infty}
{-m - 1 \choose i}\pars{-\,{z \over 2}}^{i}
\\[5mm] = &\
{1 \over 1 - z}\sum_{m = 0}^{\infty}{w^{m} \over 2^{m + 1}}
\pars{1 - {z \over 2}}^{-m - 1} =
{1 \over 1 - z}\,{1 \over 2 - z}\sum_{m = 0}^{\infty}
\pars{w \over 2 - z}^{m} =
{1 \over 1 - z}\,{1 \over 2 - z}\,{1 \over 1 - w/\pars{2 - z}}
\\[5mm] = &
\bbox[#ffe,15px]{\ds{{1 \over 1 - z}\,{1 \over 2 - w - z}}}
\end{align}
Similarly, the Generation Function of the $\ds{\color{#f00}{\texttt{second term}}}$, in \eqref{1}, is given by
$\bbox[#ffe,15px]{\ds{{1 \over 1 - w}\,{1 \over 2 - z - w}}}$.

Then,
\begin{align}
&\sum_{m, n = 0}^{\infty}\mrm{F}\pars{m,n}w^{m}z^{n} =
{1 \over 1 - z}\,{1 \over 2 - w - z} + {1 \over 1 - w}\,{1 \over 2 - z - w} =
{1 \over \pars{1 - z}\pars{1 - w}}
\end{align}

$$
\sum_{m, n = 0}^{\infty}\mrm{F}\pars{m,n}w^{m}z^{n} =
{1 \over \pars{1 - z}}\,{1 \over \pars{1 - w}}
\implies 
\bbox[15px,#ffe,border:1px dotted navy]{\ds{\large{\mrm{F}\pars{m,n} = 1}}}
$$
