Question

Determine o valor de n nas equações:
a) (n+2)! + (n+1)!=15n!
b) (n!)2-25n! +24=0

Answer 1

Olá George,


$\mathsf{A- ~(x \geq -1)}\\\mathsf{}\\\mathsf{ (n+2)! + (n+1)!=15n! \Rightarrow \dfrac{(n+2)!}{n!}+\dfrac{(n+1)!}{n!}=15}\\\\=\\\\\mathsf{\dfrac{(n+2)\cdot(n+1)\cdot\diagup\!\!\!\!n!}{\diagup\!\!\!\!n!}+\dfrac{(n+1)\cdot \diagup\!\!\!\!n!}{\diagup\!\!\!\!n!}=15}\\\\=\\\\\mathsf{(n^2+n+2n+2)+(n+1)=15\Rightarrow n^2+4n+3=15}\\\\=\\\\\mathsf{n^2+4n-12=0}\\\\\mathsf{n+n'=-4}\\\mathsf{n\cdot n'=-12}\\\\\mathsf{n=2}\\\mathsf{n'=-6\gets N\~ao~serve,~pois~\'e~menor~que~(-1)}$

$\mathsf{B-~(n\ge 1)}\\\\\mathsf{(n!)^2-25n! +24}\\\\\mathsf{n+n'=25}\\\mathsf{n\cdot n'=24}\\\\\mathsf{n=24}\\\mathsf{n'=1}$

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