Sum of combinatorics sequence $\binom{n}{1} + \binom{n}{3} +\cdots+ \binom{n}{n-1}$ I need to find sum like  $$\binom{n}{1} + \binom{n}{3} +\cdots+ \binom{n}{n-1},\qquad \text{ for even } n$$
Example:
Find the sum of $$\binom{20}{1} + \binom{20}{3} +\cdots+ \binom{20}{19}=\ ?$$
 A: Hint:
Consider the difference between:
$$ \sum_{k=0}^{20}\binom{20}{k}1^k = (1+1)^{20}\quad\text{and}\quad \sum_{k=0}^{20}\binom{20}{k}(-1)^k = (1-1)^{20}.$$
A: The important binomial theorem states that
\begin{equation}
\sum_{k=0}^n\binom{n}{k} x^ky^{n-k}=(x+y)^n
\end{equation}
Setting $x=1$ and $y=1$, we have the following relation
\begin{equation}
\sum_{k=0}^n\binom{n}{k} =2^n
\end{equation}
and setting $x=-1$ and $y=1$, we have the following relation
\begin{equation}
\sum_{k=0}^n\binom{n}{k} (-1)^k=0
\end{equation}
Hence we have
\begin{equation}
\sum_{k=1}^{n/2}\binom{n}{2k-1} =\frac{1}{2}\left(\sum_{k=0}^n\binom{n}{k}-\sum_{k=0}^{n}\binom{n}{k}(-1)^k\right)=2^{n-1}.
\end{equation}
This may be easier to see with your example
\begin{align}
&\binom{20}{1}+\binom{20}{3}+\cdots+\binom{20}{19}=\\
&\frac{1}{2}\left[\color{red}{\binom{20}{0}}+\binom{20}{1}+\color{red}{\binom{20}{2}}+\cdots+\binom{20}{20}-\left(\color{red}{\binom{20}{0}}-\binom{20}{1}+\color{red}{\binom{20}{2}}+\cdots+\binom{20}{20}\right)\right]
\end{align}
A: If you know your Pascal's triangle, you know that 
$$\binom{20}{1} + \binom{20}{3} +..+ \binom{20}{19}= \sum_{k=0}^{19} \binom{19}{k}$$
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\iff}{\Leftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
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 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
\begin{align}
\color{#f00}{\sum_{k = 0}^{9}{20 \choose 2k + 1}} & =
\sum_{k = 0}^{20}{20 \choose k}\half\bracks{1 - \pars{-1}^{k}} =
\half\braces{\pars{1 + 1}^{20} - \bracks{1 + \pars{-1}}^{20}}
\\[3mm] & = \color{#f00}{2^{19}} = 524288
\end{align}
A: Hint: Use the fact that ${n \choose k} = {n-1 \choose k-1} + {n-1 \choose k}$. Since $n$ is even, let $n = 2p$ for some $p$. Then we have:
$$\sum_{r\ \text{ odd}}{n \choose r} = \sum_{i=1}^p{2p \choose 2i-1} = \sum_{i=1}^p{2p-1 \choose 2i-2}+{2p-1 \choose 2i-1} = \sum_{i=0}^{2p-1}{2p-1 \choose i}$$
This may be easier to see with a specific example:
$${20 \choose 1} + {20 \choose 3} + \ldots + {20 \choose 19} = \left[{19 \choose 0} + {19 \choose 1}\right] + \left[{19 \choose 2} + {19 \choose 3}\right] + \left[{19 \choose 18} + {19 \choose 19}\right]$$
