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 $$\binom{20}{1} + \binom{20}{3} +\cdots+ \binom{20}{19}=\ ?$$

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}
