[tex]\frac{(n+2)!}{n!}=30\\\frac{(n+2)*(n+2-1)*(n+2-2)!}{n!}=\frac{(n+2)*(n+1)*(n)!}{n!}=\frac{(n+2)*(n+1)*1}{1}=(n+2)(n+1) = n^{2} + 3n +2[/tex]
n^2 + 3n + 2 = 30
n^2 +3n - 28 = 0
Δ = 9 - 4*1*(-28) = 9 + 112 = 121
n₁ = -3 + 11 / 2 = 8 / 2 = 4
n₂ = -3 - 11 /2 = -14 / 2 = -7 (nu convine)
Rezulta din ultimele doua relatii ca n = 4.
