Zadan je niz (an)n∈N0 takav da je a0=a, a1=b, gdje su a,b∈R, i
an=an−1+an−2,n≥2.
Odredite an2−an−1an+1.
Rješenje
Uvedimo oznaku An=an2−an−1an+1. Tada za n∈N vrijedi
An+An+1=an2−an−1an+1+an+12−anan+2=an2−an−1an+1+an+12−an(an+an+1)=an2−an−1an+1+an+12−an2−anan+1=an+12−an+1(an−1+an)=an+12−an+12=0.
Kako je An+An+1=0, to slijedi
A1A2A3An=a12−a0a2=b2−a(a+b)=b2−a2−ab,=−A1,=−A2=A1,⋮=an2−an−1an+1=(−1)n−1(b2−a2−ab).