TITLE: Silly doubt proving $2^{n-1}={n\choose 0}+ {n\choose 2}+ {n\choose 4}+ \dots $?
QUESTION [6 upvotes]: I have to prove that 
$$2^{n-1}={n\choose 0}+ {n\choose 2}+ {n\choose 4}+ \dots $$
I checked the case where $n-1$ is even but I am a little confused when $n-1$ is odd:
$$2^{n-1}={n-1\choose 0}+{n-1\choose 1}+{n-1\choose 2}+\dots +{n-1 \choose n-1}$$
I know we can group the terms pairwise with the identity ${n\choose k}+{n\choose k+1}={n+1 \choose k}$ and obtain:
$$2^{n-1}={n\choose 0}+{n\choose 2}+{n\choose 4}+\dots +{n-1 \choose n-2 } +{n-1 \choose n-1}$$
But when I do the last pair I would obtain:
$${n-1 \choose n-2 } +{n-1 \choose n-1}={n \choose n-2}$$
And the exercise says the last term should be $\displaystyle {n \choose n-1}$ or $\displaystyle  {n\choose n}$. I might be missing something truly silly, but I'm not seeing at the moment.

REPLY [2 votes]: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
\sum_{k = 0}^{\infty}{n \choose 2k} & =
\sum_{k = 0}^{\infty}{n \choose k}{1 + \pars{-1}^{k} \over 2} =
{1 \over 2}\sum_{k = 0}^{\infty}{n \choose k} + {1 \over 2}\sum_{k = 0}^{\infty}{n \choose k}\pars{-1}^{k}
\\[5mm] & =
{1 \over 2}\,\pars{1 + 1}^{\,n} + {1 \over 2}\,\bracks{1 + \pars{-1}}^{\,n} =
\bbx{\ds{2^{n - 1} + {1 \over 2}\,\delta_{n,0}}}
\end{align}