Suma de coeficientes binomiales sobre$(r+2)$

Question

ps

Intento:

$$\sum^{50}_{r=0}(-1)^r \dfrac{\dbinom {50}r}{r+2}= ?$

Integrando ambos lados y luego colocando límites 0 a 1:

$(1-x)^{50}= \sum (-1)^r \dbinom{50}r x^{n-r}$

Entonces la respuesta debería ser$\dfrac{1}{51}= \displaystyle \sum_{r=0}^{50}\dfrac{(-1)^r \dbinom{50}r}{52-r}$ pero la respuesta dada es$\dfrac{1}{51}$

¿Dónde me he equivocado?

Editar:

Entendí mi error del comentario. Ahora tengo: $1/(51\times 52)$

¿Es posible completarlo desde aquí ?

Answer 1

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove armada]{\displaystyle{#1}}\,} \newcommand{\llaves}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\parcial #3^{#1}}} \newcommand{\raíz}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \sum_{r = 0}^{50}\pars{-1}^{r}\,{{50 \choose r} \over r + 2} & = \sum_{r = 0}^{50}\pars{-1}^{r}\,{50 \choose r}\int_{0}^{1}t^{r + 1}\,\dd t = \int_{0}^{1}t\sum_{r = 0}^{50}{50 \choose r}\pars{-t}^{r}\,\dd t = \int_{0}^{1}t\,\pars{1 - t}^{50}\,\dd t \\[5mm] & = {\Gamma\pars{2}\Gamma\pars{51} \over \Gamma\pars{53}} = {1!\,50! \over 52!} = {1 \over 52 \times 51} = \bbx{1 \over 2652} \end{align}

Answer 2

Usted tiene que integrar no $(1-x)^{50}$ pero $x(1-x)^{50}$: $$ (1-x)^{50}= \sum_{i\ge0} (-1)^r \dbinom{50}r x^{r}\Rightarrow\\ x(1-x)^{50}=x\sum_{i\ge0} (-1)^r\dbinom{50}r x^{r}=\sum_{i\ge0} (-1)^r\dbinom{50}r x^{i+1}\Rightarrow\\ \int_0^1 x(1-x)^{50}dx=\left[\sum_{i\ge0} (-1)^r\dbinom{50}r\frac{x^{r+2}}{r+2}\right]_0^1=\sum_{i\ge0} (-1)^r\dbinom{50}r\frac{1}{r+2}. $$

La integral en el lado izquierdo puede ser calculada por partes: $$ \int_0^1 x(1-x)^{50}dx= \left[-\frac{x(1-x)^{51}}{51}\a la derecha]_0^1+\frac{1}{51}\int_0^1 (1-x)^{51}dx=\frac{1}{51\cdot52}. $$

Answer 3

La expresión de OP es un caso especial de una fórmula para el coeficiente binomial recíproco que usa la función Beta

\begin{align*} \color{blue}{\binom{n}{k}^{-1}}&\color{blue}{=(n+1)\int_0^1 z^k(1-z)^{n-k}\,dz}\\ &=(n+1)\int_0^1z^k\sum_{r=0}^{n-k}\binom{n-k}{r}(-z)^r\,dz\\ &=(n+1)\sum_{r=0}^{n-k}\binom{n-k}{r}(-1)^r\int_0^1z^{k+r}\,dz\\ &\,\,\color{blue}{=(n+1)\sum_{r=0}^{n-k}\binom{n-k}{r}\frac{(-1)^r}{k+r+1}}\tag{1} \end{align*}

Obtenemos de (1) \begin{align*} \sum_{r=0}^{n-k}\binom{n-k}{r}\frac{(-1)^r}{k+r+1}=\frac{1}{(n+1)\binom{n}{k}}\tag{2} \end {align *}

Poniendo$n=51$ y$k=1$ en (2) obtenemos la fórmula de OP \begin{align*} \color{blue}{\sum_{r=0}^{50}\binom{50}{r}\frac{(-1)^r}{r+2}}=\frac{1}{52\binom{51}{1}}\color{blue}{=\frac{1}{52\cdot 51}} \end {align *}