mathematical induction

U(n)=1/(√8)*(αn – βn)
V(n)=1/(√8)*(αn + βn)
a) Since x2 – 2x – 1 = 0
α = 1 + √2
β = 1 – √2
α2 – 2α – 1 = 0 OR α2 = 2α + 1
β2 – 2β – 1 = 0 OR β2 = 2β + 1
2U(n+1) + U(n)
= 2/(√8)*(αn+1 – βn+1) + 1/(√8)*(αn – βn)
=1/(√8)*[αn(2α + 1) – βn(2β + 1))
=1/(√8)*[αn(α2) – βn(β2)]
=1/(√8)*(αn+2 – βn+2)
=U(n+2)
2V(n+1) + V(n)
= 2/(√8)*(αn+1 + βn+1) + 1/(√8)*(αn + βn)
=1/(√8)*[αn(2α + 1) + βn(2β + 1))
=1/(√8)*[αn(α2) + βn(β2)]
=1/(√8)*(αn+2 + βn+2)
=V(n+2)
bi)
U(1) = 1/(√8)*(α – β) = 1/(√8)*(2√2) = 1
U(2) = 1/(√8)*(α2 – β2) = 1/(√8)*(α + β)(α – β) = 1/(√8)*(2)(2√2) = 2
ii)U(n) and U(n+1) are integers then U(n+2) = 2U(n+1)+U(n) is also an integer
iii) Let P(n) : U(n) is an integer
P(1) is true since U(1) = 1
P(2) is true since U(2) = 2
Assume P(k) and P(k+1) are true, ie,
U(k) is an integer and U(k+1) is an integer.
From (ii), U(k+2) is also an integer.
Thus by the principle of mathematical induction, P(n) is true for all positive integers.
c) Since V(1) = 1/(√8)*(α + β) =√2/2, V(n) is not always an integer.