VITEEE 2013 Paper

Multiplying x2 - ax + b = 0 by xn−1, we get
xn+1 - axn + bxn−1 = 0 ...(i)
a, p are roots of x2 - ax + b = 0, therefore they will satisfy (i).
Also, αn+1 - aαn + bαn−1 = 0 ...(ii)
and βn+1 - aβn + bβn−1 = 0 ...(iii)
On adding eqs. (ii) and (iii), we get
(αn+1+βn+1) - a(αn+βn) + b(αn−1+βn−1) = 0
⇒ Vn+1 - aVn+bVn−1 = 0 (since n+βn=Vn)
Vn+1 = aVn−bVn−1