Question

determine o valor de n na equaçao An,3= n³-40​

Answer 1

$A_{n,3}~=~n^3-40\\\\\\\frac{n!}{(n-3)!}~=~n^3-40\\\\\\\frac{n~.~(n-1)~.~(n-2)~.~(n-3)!}{(n-3)!}~=~n^3-40\\\\\\n~.~(n-1)~.~(n-2)~=~n^3-40\\\\\\n~.~(n^2-3n+2)~=~n^3-40\\\\\\n^3-3n^2+2n~=~n^3-40\\\\\\3n^2-2n-40~=~0\\\\\\Aplicando~Bhaskara\\\\\\\Delta~=~(-2)^2-4.3.(-40)~=~4+480~=~\boxed{484}\\\\\\n'~=~\frac{2+\sqrt{484}}{2~.~3}~=~\frac{2+22}{6}~=~\boxed{4}\\\\\\n''~=~\frac{2-\sqrt{484}}{2~.~3}~=~\frac{2-22}{6}~=~\boxed{-\frac{10}{3}}$

Temos 2 possibilidades para "n", no entanto precisamos lembrar que não podemos ter fatoriais nem de numeros fracionarios, nem de numeros negativos e, sendo assim, n'' deve ser descartado.

Concluímos então que "n" vale 4.