Question

como se resolve isso?

[n!+(n-1)!] / [(n+1)!]=1/6

Answer 1

Olá!

$\\ \mathsf{\frac{\left [n! + (n - 1)! \right ]}{\left [(n + 1)! \right ]} = \frac{1}{6}} \\\\\\ \mathsf{\frac{n! + (n - 1)!}{(n + 1)!} = \frac{1}{6}} \\\\\\ \mathsf{\frac{n \cdot (n - 1)! + (n - 1)!}{(n + 1) \cdot n \cdot (n - 1)!} = \frac{1}{6}} \\\\\\ \mathsf{\frac{(n - 1)! \left [ n + 1 \right ]}{(n + 1) \cdot n \cdot (n - 1)!} = \frac{1}{6}}$

$\\ \mathsf{\frac{(n + 1) \cdot (n - 1)!}{(n + 1) \cdot n \cdot (n - 1)!} = \frac{1}{6}, \qquad n \neq 1.} \\\\\\ \mathsf{\frac{1}{n} = \frac{1}{6}} \\\\ \boxed{\mathsf{n = 6}}$