Question

Se \frac{\left ( n+1 \right )!+n!}{\left ( n+2 \right )! }=\frac{1}{9}, então n é igual a: (0,2)

13

11

9

8

6

Answer 1

Resposta:

$\textsf{Leia abaixo}$

Explicação passo a passo:

$\mathsf{\dfrac{(n + 1)! + n!}{(n + 2)!} = \dfrac{1}{9}}$

$\mathsf{\dfrac{n!(n + 1) + n!}{n!(n + 1).(n + 2)} = \dfrac{1}{9}}$

$\mathsf{\dfrac{(n + 1) + 1}{(n + 1).(n + 2)} = \dfrac{1}{9}}$

$\mathsf{\dfrac{n + 2}{n^2 + 3n + 2} = \dfrac{1}{9}}$

$\mathsf{9n + 18 = n^2 + 3n + 2}$

$\mathsf{n^2 - 6n - 16 = 0}$

$\mathsf{\Delta = b^2 - 4.a.c}$

$\mathsf{\Delta = (-6)^2 - 4.1.(-16)}$

$\mathsf{\Delta = 36 + 64}$

$\mathsf{\Delta = 100}$

$\mathsf{n = \dfrac{-b \pm \sqrt{\Delta}}{2a} = \dfrac{6 \pm \sqrt{100}}{2} \rightarrow \begin{cases}\mathsf{n' = \dfrac{6 + 10}{2} = \dfrac{16}{2} = 8}\\\\\mathsf{n'' = \dfrac{6 - 10}{2} = -\dfrac{4}{2} = -2}\end{cases}}$

$\boxed{\boxed{\mathsf{S = \{8\}}}}\leftarrow\textsf{letra D}$