Question

18. Determine o conjunto solução da equação (n+2)! + (n+1)! = 24 n!​

Answer 1

Resposta:

$\textsf{Leia abaixo}$

Explicação passo a passo:

$\mathsf{(n + 2)! + (n + 1)! = 24n!}$

$\mathsf{(n + 2).(n + 1).n! + (n + 1).n! = 24n!}$

$\mathsf{(n + 2).(n + 1) + (n + 1) = 24}$

$\mathsf{n^2 + n + 2n + 2 + n + 1 = 24}$

$\mathsf{n^2 + 4n + 3 = 24}$

$\mathsf{n^2 + 4n + 3 - 24 = 0}$

$\mathsf{n^2 + 4n - 21 = 0}$

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

$\mathsf{\Delta = 4^2 - 4.1.(-21)}$

$\mathsf{\Delta = 16 + 84}$

$\mathsf{\Delta = 100}$

$\mathsf{n = \dfrac{-b \pm \sqrt{\Delta}}{2a} = \dfrac{-4 \pm \sqrt{100}}{2} \rightarrow \begin{cases}\mathsf{n' = \dfrac{-4 + 10}{2} = \dfrac{6}{2} = 3}\\\\\mathsf{n'' = \dfrac{-4 - 10}{2} = \dfrac{-14}{2} = -7}\end{cases}}$

$\boxed{\boxed{\mathsf{S = \{3\}}}}$