Question

B) (n+1)!+n!=3(n-1)!

Answer 1

$(n+1)!+n!=3(n-1)!\\\\ (n+1)n(n-1)!+n(n-1)!=3(n-1)!\\\\ ~[(n+1)n+n](n-1)!=3(n-1)!\\\\ ~\dfrac{[(n+1)n+n](n-1)!}{(n-1)!}=\dfrac{3(n-1)!}{(n-1)!}\\\\ (n+1)n+n=3\\\\ n^2+2n-3=0\\\\\\ \Delta=b^2-4ac\\\\ \Delta=2^2-4\cdot1\cdot(-3)\\\\ \Delta=4+12=16=4^2\\\\\\ n=\dfrac{-b\pm\sqrt{\Delta}}{2a}=\dfrac{-2\pm\sqrt{4^2}}{2\cdot1}=\dfrac{-2\pm4}{2}=-1\pm2\\\\ n_1=-1-2~~ou~~n_2=-1+2\\\\ n_1=-3~~~~~~~ou~~ n_2=1$

Como estamos tirando o fatorial de n na equação, temos que n é um número positivo. Logo, a única resposta que nos serve é: $\boxed{n=1}$.