Identidad combinatoria dura: $\sum_{l=0}^p(-1)^l\binom{2l}{l}\binom{k}{p-l}\binom{2k+2l-2p}{k+l-p}^{-1}=4^p\binom{k-1}{p}\binom{2k}{k}^{-1}$

Question

¿Alguien tiene alguna idea para demostrarlo? $$\sum_{\ell=0}^{p}(-1)^{\ell}\dfrac{\binom{2\ell}{\ell}\binom{k}{p-\ell}}{\binom{2k+2\ell-2p}{k+\ell-p}} = \dfrac{4^p\binom{k-1}{p}}{\binom{2k}{k}}$$ es cierto para todos los $k \in \mathbb{N}, p \in \mathbb{N} \cup \{0\}$ . Utilizamos la convención de que, si $p>k-1$ entonces $\binom{k-1}{p}=0$

Answer 1

Obtenemos \begin {align*} \color {Azul}{ \sum_ {l=0}^p}& \color {azul}{(-1)^l \binom {2l}{l} \binom {k}{p-l} \binom {2k+2l-2p}{k+l-p}^{-1}} \\ &= \sum_ {l=0}^p(-1)^{p-l} \binom {2p-2l}{p-l} \binom {k}{l} \binom {2k-2l}{k-l}^{-1} \tag {1} \\ &=4^{p-k} \sum_ {l=0}^p \binom {- \frac {1}{2}}{p-l} \binom {k}{l} \binom {k-l- \frac {1}{2}}{k-l}^{-1} \tag {2} \\ &=4^{p-k} \binom {k- \frac {1}{2}}{k}^{-1} \sum_ {l=0}^p \binom {- \frac {1}{2}}{p-l} \binom {k- \frac {1}{2}}{l} \tag {3} \\ &=4^{p-k} \binom {k- \frac {1}{2}}{k}^{-1} \binom {k-1}{p} \tag {4} \\ & \color {azul}{=4^p \binom {k-1}{p} \binom {2k}{k}^{-1}} \tag {5} \end {align*} y la afirmación sigue.

Comentario:

• En (1) cambiamos el orden de la suma $l\rightarrow p-l$ .

• En (2) utilizamos las identidades binomiales $$(-4)^n\binom{-\frac{1}{2}}{n}=\color{green}{\binom{2n}{n}}= 4^n\binom{n-\frac{1}{2}}{n}$$

• En (3) utilizamos la identidad binomial (la esencia ) \begin {align*} \binom {k- \frac {1}{2}}{l} \binom {k- \frac {1}{2}}{k}^{-1} &= \frac { \left (k- \frac {1}{2} \right )^{ \underline y que no se puede hacer nada. \cdot\frac {k!}{ \left (k- \frac {1}{2} \right )^{ \underline {k}}} \\ &= \frac {k!}{l!} \cdot\frac {1}{ \left (k-l- \frac {1}{2} \right )^{ \underline {k-l}} \\ &= \frac {k!}{l!(k-l)!} \cdot\frac {(k-l)!}{ \left (k-l- \frac {1}{2} \right )^{ \underline {k-l}} \\ &= \binom {k}{l} \binom {k-l- \frac {1}{2}}{k-l}^{-1} \\ \end {align*} con $n^{\underline{k}}=n(n-1)\cdots(n-k+1)$ el factorial descendente.

• En (4) aplicamos la Identidad Chu-Vandermonde .

• En (5) utilizamos la primera identidad establecida en (2).

Answer 2

Adoptemos la definición del Coeficiente binomial mediante el factorial descendente .

$$ \left( \matrix{ r \cr m \cr} \right) = \left\{ {\matrix{ {{{r^{\,\underline {\,m\,} } } \over {m!}}} & {\left| {\;0 \le m \in \mathbb Z} \right.} \cr 0 & {\left| {\;\neg \left( {0 \le m \in \mathbb Z} \right)} \right.} \cr } } \right. $$ con $r$ real o incluso compleja.

Ahora bien, el coeficiente binomial en el denominador de la suma deberá ser no nulo, es decir, deberá ser $0 \le k-p$ con la definición anterior. Así que pongamos $m=k-p$ y reescribir la identidad propuesta como $$ \bbox[lightyellow] { \sum\limits_{\left( {0\, \le } \right)\,l\,\left( { \le \,p} \right)} {\left( { - 1} \right)^{\,l} {{\left( \matrix{ 2l \cr l \cr} \right)\left( \matrix{ m + p \cr p - l \cr} \right)} \over {\left( \matrix{ 2l + 2m \cr m + l \cr} \right)}}} = 4^{\,p} {{\left( \matrix{ m + p - 1 \cr p \cr} \right)} \over {\left( \matrix{ 2m + 2p \cr m + p \cr} \right)}}\quad \left| {\;0 \le m,p \in \mathbb Z} \right. } \tag{1}$$ Nótese que los límites de la suma pueden omitirse ya que están implícitos en los binomios.

Con la definición adoptada, el binomio central puede escribirse como $$ \left( \matrix{ 2n \cr n \cr} \right) = 2^{\,2\,n} \left( \matrix{ n - 1/2 \cr n \cr} \right) = \left( { - 1} \right)^{\,n} 2^{\,2\,n} \left( \matrix{ - 1/2 \cr n \cr} \right) = \left( { - 4} \right)^{\,n} \left( \matrix{ - 1/2 \cr n \cr} \right) $$

Recordemos entonces que, para el Factorial Ascendente y Descendente tenemos $$ \eqalign{ & 1 = x^{\,\underline {\,0\,} } = x^{\,\underline {\,m\,} } \left( {x - m} \right)^{\,\underline {\, - m\,} } \cr & \left( {x + 1} \right)^{\,\overline {\, - m\,} } = \left( {x - m} \right)^{\,\underline {\, - m\,} } = {1 \over {x^{\,\underline {\,m\,} } }}\quad x^{\,\underline {\, - m\,} } = {1 \over {\left( {x + m} \right)^{\,\underline {\,m\,} } }} = {1 \over {\left( {x + 1} \right)^{\,\overline {\,m\,} } }} \cr} $$

Con esta premisa, los sumandos en el LHS se pueden escribir como $$ \bbox[lightyellow] { \eqalign{ & \left( { - 1} \right)^{\,l} {{\left( \matrix{ 2l \cr l \cr} \right)\left( \matrix{ m + p \cr p - l \cr} \right)} \over {\left( \matrix{ 2l + 2m \cr m + l \cr} \right)}} = {{4^{\,l} \left( \matrix{ - 1/2 \cr l \cr} \right)\left( \matrix{ m + p \cr m + l \cr} \right)} \over {4^{\,m + l} \left( \matrix{ m + l - 1/2 \cr m + l \cr} \right)}} = \cr & = {1 \over {4^{\,m} l!}}{{\left( { - 1/2} \right)^{\,\underline {\,l\,} } \left( {m + p} \right)^{\,\underline {\,m + l\,} } } \over {\left( {m + l - 1/2} \right)^{\,\underline {\,m + l\,} } }} \cr} } \tag{2}$$

Y podemos reescribir la RHS como $$ \bbox[lightyellow] { \eqalign{ & RHS = 4^{\,p} {{\left( \matrix{ m + p - 1 \cr p \cr} \right)} \over {\left( \matrix{ 2m + 2p \cr m + p \cr} \right)}} = 4^{\,p} {{\left( \matrix{ m + p - 1 \cr p \cr} \right)} \over {4^{\,m + p} \left( \matrix{ m + p - 1/2 \cr m + p \cr} \right)}} = \cr & = {1 \over {4^{\,m} }}{{\sum\limits_{\left( {0\, \le } \right)\,l\,\left( { \le \,p} \right)} {\left( \matrix{ - 1/2 \cr l \cr} \right)\left( \matrix{ m + p - 1/2 \cr p - l \cr} \right)} } \over {\left( \matrix{ m + p - 1/2 \cr m + p \cr} \right)}} \cr} } \tag{3}$$

y podemos demostrar que cada sumando de la izquierda corresponde a un sumando de la derecha $$ \bbox[lightyellow] { \eqalign{ & {1 \over {4^{\,m} l!}}{{\left( { - 1/2} \right)^{\,\underline {\,l\,} } \left( {m + p} \right)^{\,\underline {\,m + l\,} } } \over {\left( {m + l - 1/2} \right)^{\,\underline {\,m + l\,} } }} = {1 \over {4^{\,m} }}{{\left( \matrix{ - 1/2 \cr l \cr} \right)\left( \matrix{ m + p - 1/2 \cr p - l \cr} \right)} \over {\left( \matrix{ m + p - 1/2 \cr m + p \cr} \right)}} \cr & \quad \quad \Downarrow \cr & {{\left( {m + p} \right)^{\,\underline {\,m + l\,} } } \over {\left( {m + l - 1/2} \right)^{\,\underline {\,m + l\,} } }} = {{\left( {m + p} \right)!\left( {m + p - 1/2} \right)^{\,\underline {\,p - l\,} } } \over {\left( {p - l} \right)!\left( {m + p - 1/2} \right)^{\,\underline {\,m + p\,} } }} \cr & \quad \quad \Downarrow \cr & {{\left( {m + p} \right)^{\,\underline {\,m + l\,} } } \over {\left( {m + l} \right)!}}{1 \over {\left( {m + l - 1/2} \right)^{\,\underline {\,m + l\,} } }} = {{\left( {m + p} \right)!} \over {\left( {p - l} \right)!\left( {m + l} \right)!}}{{\left( {m + p - 1/2} \right)^{\,\underline {\,p - l\,} } } \over {\left( {m + p - 1/2} \right)^{\,\underline {\,m + p\,} } }} \cr & \quad \quad \Downarrow \cr & {1 \over {\left( {m + l - 1/2} \right)^{\,\underline {\,m + l\,} } }} = {{\left( {m + p - 1/2} \right)^{\,\underline {\,p - l\,} } } \over {\left( {m + p - 1/2} \right)^{\,\underline {\,m + p\,} } }} \cr & \quad \quad \Downarrow \cr & \left( {m + p - 1/2} \right)^{\,\underline {\,m + p\,} } = \left( {m + p - 1/2} \right)^{\,\underline {\,\,p - l\, + m + l\,} } = \cr & = \left( {m + p - 1/2} \right)^{\,\underline {\,p - l\,} } \left( {m + l - 1/2} \right)^{\,\underline {\,m + l\,} } \cr} } \tag{4}$$