This formula in sum notation ${n \choose 0} + {n \choose 1} + ... + {n \choose (n-1)/2}$ For $n \in \mathbb N$, where $n$ are odd numbers only. I tried different combination, but I was not able to find out the correct formula.
Yeah, I know the result is $2^{n-1}$, but I need sum notation in order to be able to prove it.
Thanks
 A: I suppose the question is framed so that you aren't given $2^{n-1}$ but rather have to deduce it, but I still think it'd help to look at a combinatorial answer. $\sum_{k=0}^{\frac{n-1}{2}} {n\choose k}$ counts the number of subsets of $\{1,2...n\}$ (which we'll call $[n]$) of size at most $\frac{n-1}{2}$- let's call this collection of subsets $\mathcal{A}$. $2^{n-1}$ counts the number of subsets of $[n-1]$- let's call this collection of subsets $\mathcal{B}$. Let $A\in \mathcal{A}$. If $A$ doesn't contain $n$, then it is a subset of $[n-1]$- this gives us all subsets of $[n-1]$ of size less than $\frac{n-1}{2}$. Now that we have an obvious bijection between sets in $\mathcal{A}$ that don't contain $n$ and the sets in $\mathcal{B}$ of size at most $\frac{n-1}{2}$, we just need a bijection between sets in $\mathcal{A}$ that contain $n$ and sets in $\mathcal{B}$ of size greater than $\frac{n-1}{2}$.
Consider $A\in\mathcal{A},n\in A$. Now consider $[n-1]-(A-\{n\})$. In other words, we use the presence of $n$ to tell us to look at the complement of $A-\{n\}$ in $[n-1]$. Since $|A|\leq\frac{n-1}{2}$, $|A-\{n\}|<\frac{n-1}{2}$ and so $|[n-1]-(A-\{n\})|>\frac{n-1}{2}$, which gives us exactly the kind of set in $\mathcal{B}$ we desired. We can now look at the inverse function by considering a set $B$ in $\mathcal{B}$ of size greater than $\frac{n-1}{2}$. Then $([n-1]-B)\cup\{n\}$ gives a subset of $[n]$ with size at most $\frac{n-1}{2}$ that contains $n$, and so we're done.
In other words, we have a function $f:\mathcal{A}\to\mathcal{B}$ defined by
$f(A)=$
\begin{cases} 
      A & n\notin A \\
      [n-1]-(A-\{n\}) & n\in A
\end{cases}
and its inverse $g:\mathcal{B}\to\mathcal{A}$ defined by $g(B)=$
\begin{cases} 
      B & |B|\leq\frac{n-1}{2}\\
      [n-1]-B & |B|>\frac{n-1}{2}
\end{cases}
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
\left.\sum_{k = 0}^{\pars{n - 1}/2}{n \choose k}\right\vert_{\ n\ \mrm{odd}} & =
\sum_{k = 0}^{n}{n \choose k}\bracks{k \leq {n - 1 \over 2}}
\\[5mm] & =
{1 \over 2}
\sum_{k = 0}^{n}\braces{{n \choose k}\bracks{k \leq {n - 1 \over 2}} +
{n \choose n - k}\bracks{n - k \leq {n - 1 \over 2}}}
\\[5mm] & =
{1 \over 2}
\sum_{k = 0}^{n}{n \choose k}\braces{\bracks{k \leq {n - 1 \over 2}} +
\bracks{k \geq {n + 1 \over 2}}} =
{1 \over 2}\sum_{k = 0}^{n}{n \choose k} = {1 \over 2}\,2^{n}
\\[5mm] & = \bbx{\ds{2^{n -1}}}
\end{align}
A: What about
$$\binom n0+\binom n1+\cdots+\binom n{\frac{n-1}{2}}=\sum_{k=0}^{\frac{n-1}{2}}\binom nk$$
