Show that $\sum_{k=0}^n\binom{3n}{3k}=\frac{8^n+2(-1)^n}{3}$

Solution 1:

$$f(x)=\sum_0^{3n}{3n\choose r}x^r=(1+x)^{3n}$$ Now let $a,b$ be the nonreal third roots of 1, and evaluate $$f(1)+f(a)+f(b)$$

Solution 2:

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $$ \mbox{Note that}\quad\sum_{k = 0}^{n}{3n \choose 3k} =\sum_{k = 0}^{\infty}{3n \choose 3k} $$

\begin{align} &\color{#c00000}{\sum_{k = 0}^{n}{3n \choose 3k}}= \sum_{k = 0}^{\infty}\oint_{\verts{z}\ =\ a\ >\ 1} {\pars{1 + z}^{3n} \over z^{3k + 1}}\,{\dd z \over 2\pi\ic} =\oint_{\verts{z}\ =\ a\ >\ 1}{\pars{1 + z}^{3n} \over z} \sum_{k = 0}^{\infty}\pars{1 \over z^{3}}^{k}\,{\dd z \over 2\pi\ic} \\[5mm]&=\oint_{\verts{z}\ =\ a\ >\ 1}{\pars{1 + z}^{3n} \over z} {1 \over 1 - 1/z^{3}}\,{\dd z \over 2\pi\ic} =\oint_{\verts{z}\ =\ a\ >\ 1} {z^{2}\pars{1 + z}^{3n} \over z^{3} - 1}\,{\dd z \over 2\pi\ic} \end{align}

The integrand has three simple poles inside the contour: $\quad\ds{z_{m} \equiv \expo{2m\pi\ic/3}\,,\quad m = -1,0,1}$: \begin{align} &\color{#c00000}{\sum_{k = 0}^{n}{3n \choose 3k}}= \sum_{m = -1}^{1}\lim_{z \to z_{m}} \bracks{\pars{z - z_{m}}\,{z^{2}\pars{1 + z}^{3n} \over z^{3} - 1}} =\sum_{m = -1}^{1}{z_{m}^{2}\pars{1 + z_{m}}^{3n} \over 3z_{m}^{2}} \\[5mm]&={1 \over 3}\sum_{m = -1}^{1}\pars{1 + \expo{2m\pi\ic/3}}^{3n} ={1 \over 3}\sum_{m = -1}^{1}\expo{mn\pi\ic} \pars{\expo{-m\pi\ic/3} + \expo{m\pi\ic/3}}^{3n} \\[5mm]&={8^{n} \over 3}\sum_{m = -1}^{1} \pars{-1}^{mn}\cos^{3n}\pars{m\,{\pi \over 3}} \\[5mm]&={8^{n} \over 3}\bracks{\pars{-1}^{-n}\cos^{3n}\pars{-\,{\pi \over 3}} + 1 + \pars{-1}^{n}\cos^{3n}\pars{\pi \over 3}} ={8^{n} \over 3}\bracks{1 + 2\pars{-1}^{n}\pars{\half}^{3n}} \\[5mm]&={8^{n} \over 3}\bracks{1 + {2\pars{-1}^{n} \over 8^{n}}} \end{align}

$$ \color{#66f}{\large\sum_{k = 0}^{n}{3n \choose 3k} ={8^{n} + 2\pars{-1}^{n} \over 3}} $$