(n+2)!+(n+1)!=35n!
n!(n+1)(n+2)+n!(n+1)=35n!
Inmultind toata ecuatia cu 1/n!, avem:
(n+1)(n+2)+(n+1)=35
(n+1)(n+2+1)=35
(n+1)(n+3)=35
n²+3n+n+3-35=0
n²+4n-32=0
Δ=16-4*(-32)=16+128=144
√Δ=12
Dar cum n apartine N, luam doar valoarea pozitiva a lui n:
n=(-4+12)/2=8/2=4
