Problema do mês :: Problem of the month #2. (Coeficientes da série binomial :: Binomial series coefficients). Resolução :: Solution

ver/see Problema do mês Problem of the month

Problema: Admita que n=1,2,3,\dots . Seja x\ge 0 um número real, \dbinom{x}{0}=1 e

\dbinom{x}{n}=\dfrac{x\left( x-1\right) \cdots\left( x-n+1\right) }{n!}

Deduza a identidade

\dbinom{x}{n}+\dbinom{x}{n-1}=\dbinom{x+1}{n}.

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Resolução de antonio girao

[que usou a notação \dfrac{x!}{\left( x-n\right) !}=x\left( x-1\right) \cdots \left( x-n+1\right) ].

\dbinom{x}{n}=\dfrac{x!}{\left( x-n\right) !n!}

\dbinom{x}{n-1}=\dfrac{x!}{\left( x-n+1\right) !\left( n-1\right) !}

\dfrac{x!}{\left( x-n\right) !n!}+\dfrac{x!}{\left( x-n+1\right) !\left( n-1\right) !}

 =\dfrac{x!}{\left( x-n\right) !n!}+\dfrac{x!n}{\left( x-n+1\right) !\left( n-1\right) !n}

 =\dfrac{x!\left( x-n+1\right) }{\left( x-n\right) !n!\left( x-n+1\right) }+\dfrac{x!n}{\left( x-n+1\right) !\left( n-1\right) !n}

=\dfrac{x!\left( x-n+1\right) }{\left( x-n+1\right) !n!}+\dfrac{x!n}{\left( x-n+1\right) !n!}

=\dfrac{x!\left[ \left( x-n+1\right) +n\right] }{\left( x-n+1\right) !n!}

 =\dfrac{x!\left( x+1\right) }{\left( x-n+1\right) !n!}

=\dfrac{\left( x+1\right) !}{\left( x+1-n\right) !n!}

=\dbinom{x+1}{n}

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Outros: Pierre Bernard (aqui) e MathOMan (aqui).

* * *

Problem: Suppose that n=1,2,3,\dots . Let x\ge 0 be a real number, \dbinom{x}{0}=1 and

\dbinom{x}{n}=\dfrac{x\left( x-1\right) \cdots\left( x-n+1\right) }{n!}.

Derive the identity

\dbinom{x}{n}+\dbinom{x}{n-1}=\dbinom{x+1}{n}.

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Solution by antonio girao

[who used the notation \dfrac{x!}{\left( x-n\right) !}=x\left( x-1\right) \cdots \left( x-n+1\right) ].

\dbinom{x}{n}=\dfrac{x!}{\left( x-n\right) !n!}

\dbinom{x}{n-1}=\dfrac{x!}{\left( x-n+1\right) !\left( n-1\right) !}

\dfrac{x!}{\left( x-n\right) !n!}+\dfrac{x!}{\left( x-n+1\right) !\left( n-1\right) !}

 =\dfrac{x!}{\left( x-n\right) !n!}+\dfrac{x!n}{\left( x-n+1\right) !\left( n-1\right) !n}

 =\dfrac{x!\left( x-n+1\right) }{\left( x-n\right) !n!\left( x-n+1\right) }+\dfrac{x!n}{\left( x-n+1\right) !\left( n-1\right) !n}

=\dfrac{x!\left( x-n+1\right) }{\left( x-n+1\right) !n!}+\dfrac{x!n}{\left( x-n+1\right) !n!}

=\dfrac{x!\left[ \left( x-n+1\right) +n\right] }{\left( x-n+1\right) !n!}

 =\dfrac{x!\left( x+1\right) }{\left( x-n+1\right) !n!}

=\dfrac{\left( x+1\right) !}{\left( x+1-n\right) !n!}

=\dbinom{x+1}{n}

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Other solvers: Pierre Bernard (here) e MathOMan (here).

* * *

Notas:

1. Estes coeficientes são os da série binomial

(1+t)^x=\displaystyle\sum_{n=0}^{\infty}\dbinom{x}{n}t^n,

que é convergente para \left\vert t\right\vert <1.

2. O coeficiente de ordem n é um polinómio de grau n em x.

Remarks:

1. These coefficients are the binomial series ones

(1+t)^x=\displaystyle\sum_{n=0}^{\infty}\dbinom{x}{n}t^n,

which is convergent for \left\vert t\right\vert <1.

2. The coefficient of order n is a polynomial of degree n in x.

Sobre Américo Tavares

eng. electrotécnico reformado / retired electrical engineer
Esta entrada foi publicada em Matemática, Math, Problem, Problem Of The Month, Problema do mês, Problemas. ligação permanente.

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