Question

Determine n na equação , que envolve números binomiais.
$\left(\begin{array}{cc}1999\\2n-1\\\end{array}\right)=\left(\begin{array}{cc}2000\\2n-200\\\end{array}\right)-\left(\begin{array}{cc}1999\\1999-2n\\\end{array}\right)$

Answer 1

Resposta:

Explicação passo-a-passo:

Dados dois números binomiais

$\left(\begin{array}{cc}n\\p\\\end{array}\right)=\left(\begin{array}{ccc}n\\q\\\end{array}\right)$, temos:

p = q ou p + q = n

Sabemos também da relação de Stiffel, que:

$\left(\begin{array}{cc}n\\p\\\end{array}\right)+\left(\begin{array}{ccc}n\\p+1\\\end{array}\right)=\left(\begin{array}{cc}n+1\\p+1\\\end{array}\right)$

$\left(\begin{array}{cc}1999\\2n-1\\\end{array}\right)+\left(\begin{array}{cc}1999\\1999-2n\\\end{array}\right)=\left[\begin{array}{cc}2000\\2n-200\\\end{array}\right]$

O complementar de:

$\left(\begin{array}{ccc}1999\\1999-2n\\\end{array}\right)=\left(\begin{array}{cc}1999\\1999-1999+2n\\\end{array}\right)=\left(\begin{array}{cc}1999\\2n\\\end{array}\right)$

$\left(\begin{array}{cc}1999\\2n-1\\\end{array}\right)+\left(\begin{array}{cc}1999\\2n\\\end{array}\right)=\left(\begin{array}{cc}2000\\2n-200\\\end{array}\right)\\\\\left(\begin{array}{cc}2000\\2n\\\end{array}\right)=\left(\begin{array}{cc}2000\\2n-200\\\end{array}\right)$

Logo: 2n = 2n - 200 ⇒ 0n = 200, (Não existe n natural)

ou

2n + 2n - 200 = 2000

4n = 2200 ⇒ n = 1200/4

n = 550