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

The other day a friend of mine showed me this sum: $\sum_{k=0}^n\binom{3n}{3k}$. To find the explicit formula I plugged it into mathematica and got $\frac{8^n+2(-1)^n}{3}$. I am curious as to how one would arrive at this answer.

My progress so far has been limited. I have mostly been trying to see if I can somehow relate the sum to $$\sum_{k=0}^{3n}\binom{3n}{k}=8^n$$ but I'm not getting very far. I have also tried to write it out in factorial form, but that hasn't helped me much either.

How would I arrive at the explicit formula?

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math.stackexchange.com/q/918/152 – Grigory M Oct 14 '12 at 12:50

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

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It can be proved fairly straightforwardly by induction.

Let $S_0(n)=\sum_{k\ge 0}\binom{3n}{3k}$, $S_1(n)=\sum_{k\ge 0}\binom{3n}{3k+1}$, and $S_2(n)=\sum_{k\ge 0}\binom{3n}{3k+2}$; then $$S_0(n)+S_1(n)+S_2(n)=\sum_{k\ge 0}\binom{3n}k=2^{3n}=8^n\;.$$

Now

\begin{align*} S_0(n)&=\sum_{k\ge 0}\binom{3n}{3k}\\ &=\sum_{k\ge 0}\left(\binom{3n-3}{3k-3}+3\binom{3n-3}{3k-2}+3\binom{3n-3}{3k-1}+\binom{3n-3}{3k}\right)\\ &=\sum_{k\ge 0}\left(\binom{3n-3}{3k-3}+\binom{3n-3}{3k-2}+\binom{3n-3}{3k-1}\right)\\ &\qquad\qquad+\sum_{k\ge 0}\left(\binom{3n-3}{3k-2}+\binom{3n-3}{3k-1}+\binom{3n-3}{3k}\right)\\ &\qquad\qquad+\sum_{k\ge 0}\left(\binom{3n-3}{3k-2}+\binom{3n-3}{3k-1}\right)\\ &=S_0(n-1)+S_1(n-1)+S_2(n-1)\\ &\qquad\qquad+S_0(n-1)+S_1(n-1)+S_2(n-1)\\ &\qquad\qquad+S_1(n-1)+S_2(n-1)\\ &=3\cdot8^{n-1}-S_0(n-1)\\ &=3\cdot8^{n-1}-\frac{8^{n-1}+2(-1)^{n-1}}3\qquad\qquad\text{by the induction hypothesis}\\ &=\frac{8^n-2(-1)^{n-1}}3\\ &=\frac{8^n+2(-1)^n}3\;. \end{align*}

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

\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}

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

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I'll try to give a more detailed version of Gerry Myerson's hint.

If $S$ is your sum then you have $$3S=(1+1)^{3n}+(1+e^{i\frac{2\pi}3})^{3n}+(1+e^{-i\frac{2\pi}3})^{3n}.$$ (To get this observe which terms get cancelled. If you are not familiar with this way of writing complex numbers, see Wikipedia.)

Now we want to simplify $(1+e^{i\frac{2\pi}3})^{3n}+(1+e^{-i\frac{2\pi}3})^{3n}$. We notice (by a direct computation - it helps if you draw a picture) that $1+e^{i\frac{2\pi}3}=e^{i\frac\pi3}$ and $1+e^{i\frac{2\pi}3}=e^{-i\frac\pi3}$. $$(1+e^{i\frac{2\pi}3})^{3n}+(1+e^{-i\frac{2\pi}3})^{3n} = (e^{i\frac\pi3})^{3n}+(e^{-i\frac\pi3})^{3n}=e^{in\pi}+e^{-in\pi}=(e^{i\pi})^n+(e^{-i\pi})^n.$$ Since $e^{i\pi}=e^{-i\pi}=-1$, you get $$3S=2^{3n}+2(-1)^n=8^n+2(-1)^n.$$

The trick is very similar to the using $(1+1)^n+(1-1)^n$ to get the sum of even binomial coefficients, see this question: Evaluate $\binom{n}{0}+\binom{n}{2}+\binom{n}{4}+\cdots+\binom{n}{2k}+\cdots$

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