Question

determine o valor de n na igualdade (n+1)!/n!(n-2)!=2/(n-3)!? me ajudeeem por favor!

Answer 1

$\mathrm{\dfrac{(n+1)!}{n!(n-2)!}=\dfrac{2}{(n-3)!}\ \to\ \dfrac{(n+1)n!}{n!(n-2)(n-3)!}=\dfrac{2}{(n-3)!}}\\\\\\ \mathrm{\dfrac{n+1}{n-2}=2\ \to\ n+1=2(n-2)\ \to\ n+1=2n-4}\\\\\\ \mathrm{2n-n=4+1\ \to\ \boxed{\mathbf{n=5}}}$