sí, es $$\sum_{i,j,k\in \mathbb{Z}}\binom{n}{i+j}\binom{n}{j+k}\binom{n}{i+k}$ $ y nos estamos sumando sobre todas posibles tríos de números enteros. Parece bastante obvio que el resultado no es un infinito. He intentado calcular $$\sum_{i,j \in \mathbb{Z}} \binom{n}{i+j}\sum_{k \in \mathbb{Z}} \binom{n}{i+k} \binom{n}{j+k}$ $ pero no más fácil para mí. Mi intuición dice que estamos calculando algo sobre conjunto de tamaño $3n$, pero no pude conseguir cualquier idea de derecho. Agradeceria alguna ayuda en esta tarea de examen super viejo.
Respuestas
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La clave está dada en cualquier $(u,v,w) \in \mathbb{Z}^3$, uno puede encontrar $(i,j,k) \in \mathbb{Z}^3$ tal que $$\begin{cases} u &= i + j,\\ v &= j + k,\\ w &= k + i \end{casos}$$ cuando y sólo cuando $u + v + w$ es incluso. Además, el $(i,j,k)$ asociado a $(u,,v,w)$ es único, si es que existe. Esto lleva a
$$ \sum_{i,j,k \in \mathbb{Z}}\binom{n}{i+j} \binom{n}{j+k}\binom{n}{k+i} = \sum_{\substack{u,v,w \in \mathbb{Z},\\ u+v+w\text{ par}}} \binom{n}{u}\binom{n}{v}\binom{n}{w} = \sum_{\substack{0 \le u, v, w \le n\\ u+v+w\text{ par}}} \binom{n}{u}\binom{n}{v}\binom{n}{w}\\ = \frac12 \sum_{0 \le u, v, w \le n}\left(1 + (-1)^{u+v+w}\right)\binom{n}{u}\binom{n}{v}\binom{n}{w} = \frac12\left\{\left[\sum_{u=0}^n \binom{n}{u}\right)^3 + \left[\sum_{u=0}^n (-1)^u\binom{n}{u}\right)^3 \right\}\\ = \frac12\left[(1+1)^{3n} + (1-1)^{3n}\right] \stackrel{\color{blue}{[1]}}{=} \frac12 \left[2^{3n} + \begin{cases}0,& n > 0\\1,&n = 0\end{casos}\right] = \begin{cases}2^{3n-1},&n > 0\\1, &n = 0\end{casos} $$
Notas
- $\color{blue}{[1]}$ - gracias a @FelixMartin señalando el caso especial en $n = 0$.
Felix Marin
Puntos
32763
$\newcommand{\ángulos}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\llaves}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\mitad}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\parcial #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\raíz}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{equation} \mathbf{\mbox{The Question:}\quad} \sum_{j,k,\ell\ \in\ \mathbb{Z}}\,\, {n \choose j + k}{n \choose k + \ell}{n \choose j + \ell} =\ ?\tag{1} \end{equation}
\begin{equation}\mbox{Note that}\ \sum_{j,k,\ell\ \in\ \mathbb{Z}}\,\, {n \choose j + k}{n \choose k + \ell}{n \choose j + \ell}= \sum_{j,k\ \in\ \mathbb{Z}}\,\, {n \choose j + k}\ \overbrace{\sum_{\ell\ \in\ \mathbb{Z}}{n \choose k + \ell} {n \choose j + \ell}}^{\ds{\equiv\ \,\mathcal{I}}}\tag{2} \end{equation}
\begin{align} \fbox{%#%#%} & = \sum_{\ell\ \in\ \mathbb{Z}}{n \choose k + \ell}{n \choose j + \ell} = \sum_{\ell\ \in\ \mathbb{Z}}{n \choose k + \ell}{n \choose n - j - \ell} \\[4mm] & = \sum_{\ell\ \in\ \mathbb{Z}}{n \choose k + \ell}\oint_{\verts{z} = 1^{-}} {\pars{1 + z}^{n} \over z^{n - j - \ell + 1}}\,{\dd z \over 2\pi\ic} = \oint_{\verts{z} = 1^{-}} {\pars{1 + z}^{n} \over z^{n - j + 1}} \sum_{\ell\ \in\ \mathbb{Z}}{n \choose k + \ell}z^{\ell}\,{\dd z \over 2\pi\ic} \\[4mm] & = \oint_{\verts{z} = 1^{-}} {\pars{1 + z}^{n} \over z^{n - j + k + 1}} \sum_{\ell\ \in\ \mathbb{Z}}{n \choose \ell}z^{\ell}\,{\dd z \over 2\pi\ic} \\[4mm] & = \oint_{\verts{z} = 1^{-}}{\pars{1 + z}^{n} \over z^{n - j + k + 1}} \pars{1 + z}^{n}\,{\dd z \over 2\pi\ic} = {2n \choose n - j + k} = {2n \choose n + j - k} \\[4mm] & = \fbox{$\ds{\ \oint_{\verts{z} = 1^{-}} {\pars{1 + z}^{2n} \over z^{n + j - k + 1}}\,{\dd z \over 2\pi\ic}\ }$} = \fbox{%#%#%} \end{align}
El original de la sumación $\ds{\ \,\mathcal{I}\ }$ se reduce a $\ds{\ \,\mathcal{I}\ }$: \begin{align} &\color{#f00}{\sum_{j,k,\ell\ \in\ \mathbb{Z}}\,\, {n \choose j + k}{n \choose k + \ell}{n \choose j + \ell}} = \sum_{j,k\ \in\ \mathbb{Z}}{n \choose j + k}\oint_{\verts{z} = 1^{-}} {\pars{1 + z}^{2n} \over z^{n + j - k + 1}}\,{\dd z \over 2\pi\ic} \\[4mm] = &\ \sum_{j\ \in\ \mathbb{Z}}\,\,\oint_{\verts{z} = 1^{-}} {\pars{1 + z}^{2n} \over z^{n + j + 1}} \sum_{k\ \in\ \mathbb{Z}}{n \choose j + k}z^{k}\,{\dd z \over 2\pi\ic} = \sum_{j\ \in\ \mathbb{Z}}\,\,\oint_{\verts{z} = 1^{-}} {\pars{1 + z}^{2n} \over z^{n + 2j + 1}} \sum_{k\ \in\ \mathbb{Z}}{n \choose k}z^{k}\,{\dd z \over 2\pi\ic} \\[4mm] = &\ \sum_{j\ \in\ \mathbb{Z}}\,\,\oint_{\verts{z} = 1^{-}} {\pars{1 + z}^{3n} \over z^{n + 2j + 1}}\,{\dd z \over 2\pi\ic} = \sum_{j\ \in\ \mathbb{Z}}{3n \choose n + 2j} = \sum_{j\ \in\ \mathbb{Z}}{3n \choose n + j}\,{1 + \pars{-1}^{j} \over 2} \\[4mm] = &\ \half\sum_{j\ \in\ \mathbb{Z}}{3n \choose j} + \half\,\pars{-1}^{n}\sum_{j\ \in\ \mathbb{Z}}{3n \choose j}\pars{-1}^{j} = \half\,\pars{1 + 1}^{3n} + \half\,\pars{-1}^{n}\pars{1 - 1}^{3n} \\[4mm] = &\ \color{#f00}{2^{3n - 1} + \half\,\delta_{n0}} \end{align}
