Question

Se (n+1)!=10.n!, então (n-1)2ª vale:

Answer 1

$(n + 1)! = 10 \cdot n! \\ \frac{(n + 1)!}{n!} = 10 \\ \frac{(n + 1) \cdot n!}{n!} = 10 \\ n + 1 = 10 \\ n + 1 - 2 = 10 - 2 \\ n - 1 = 8 \\ {(n - 1)}^{2} = {8}^{2} \\ {(n - 1)}^{2} = 64$