Question

Sa se  rezolve in N inecuatia
$\frac{16n!}{(n-1)!} > \frac{5n!}{(2-n)!}$

Answer 1

   
$\frac{16n!}{(n-1)!}> \frac{5n!}{(n-2)!} \\ \\ \frac{16(n-1)!*n}{(n-1)!}> \frac{5(n-2)!*(n-1)*n}{(n-2)!} \\ \\ 16n>5n(n-1) \\ 16>5(n-1) \\ 5(n-1)<16 \\ \\ n-1< \frac{16}{5} \\ \\ n<\frac{16}{5}+1 \\ \\ n<\frac{21}{5} \\ \\ n < 4,2 \\ n \,\,apartine\,\,(0; 1; 2; 3; 4)$