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

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

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

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

$\bigskip$

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

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

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

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

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

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

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

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

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

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

$=\dbinom{x+1}{n}$

$\bigskip$

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

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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

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

$\bigskip$

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

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

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

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

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

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

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

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

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

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

$=\dbinom{x+1}{n}$

$\bigskip$

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 $x$.

Anúncios

## 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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