How to prove two binomial expressions are equivalent?

Question

I have two binomial expression: $S1= \sum_{k=0}^{\frac{n}{2}}{{n}\choose{2k}} $ and $ S2 =\sum_{k=0}^{\frac{n-1}{2}}{{n}\choose{2k + 1}} $. I have to prove those two expression are equal, and then find their shared value.

So far I'm trying to find a telescoping sum such as S1-S2=0 but can't do it.I'm probably missing something.

Answer 1

Hint

For natural number $n,$

$$(1+x)^n+(1-x)^n=?$$

$$(1+x)^n-(1-x)^n=?$$

Then set $x=1$

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\left\{\begin{array}{rccclcl} \ds{S_{1}} & \ds{\equiv} & \ds{\sum_{k = 0}^{\left\lfloor n/\pars{2k}\right\rfloor}{n \choose 2k}} & \ds{=} & \ds{\sum_{k = 0}^{\infty}{n \choose k}{1 + \color{red}{\pars{-1}^{k}} \over 2}} & \ds{=} & \ds{2^{n - 1}} \\[5mm] \ds{S_{2}} & \ds{\equiv} & \ds{\sum_{k = 0}^{\left\lfloor \pars{n - 1}/\pars{2k}\right\rfloor} {n \choose 2k + 1}} & \ds{=} & \ds{\sum_{k = 0}^{\infty}{n \choose k}{1 - \color{red}{\pars{-1}^{k}} \over 2}} & \ds{=} & \ds{2^{n - 1}} \end{array}\right. \end{align} The $\ds{\color{red}{\tt red\ factors}}$ don't yield any contribution when $\ds{\underline{\underline{n > 0}}}$.