Let p, q be integers and let α,β be the roots of x2−x−1=0, where α≠β. For n=0,1,2,… Let an=pαn+qβ2
Then,
an+1=an+an−1
an+2=an+1+an−1
an+1=an+1
an+1=an−1+1
It is given that α and β are roots of x2−x−1=0
∴ α+β=1 and αβ=−1
We have an=pαn+qβn
∴ an+1=pαn+1+qβn+1⇒ an+1=pαn+qβn(α+β)−αβpαn−1+qβn−1⇒ an+1=an+an−1 [∵α+β=1,αβ=−1]