Second order different equation

Let the sequence be A(i).
A(n)= B A(n-1)+C A(n-2), B,C are constants
Let h,k be the roots of equation x^2-Bx-C
Then h+k=B, hk=-C
A(n)= B A(n-1)+C A(n-2)
A(n)- BA(n-1)-C A(n-2)=0
A(n)- (h+k)A(n-1)+hk A(n-2)=0
A(n)- h A(n-1)-kA(n-1)+hk A(n-2)=0
(A(n)- h A(n-1))-k(A(n-1)-h A(n-2))=0

Let H(n)= A(n)- h A(n-1)
H(n)-k H(n-1)=0
H(n)/H(n-1)=k
By telescopic property,
H(n)/H(n-1)=k
H(n-1)/H(n-2)=k
...
...
H(2)/H(1)=k
Multiply the above equation, H(2) to H(n-1) is cancelled.
H(n)/H(1)=k^(n-1)
H(n)= H(1) k^(n-1)
Very similarly, let K(n)= A(n)- k A(n-1)
K(n)= K(1) h^(n-1)

Now we have two formulas
H(n)= H(1) k^(n-1)
K(n)= K(1) h^(n-1)

A(n)-h A(n-1)= H(1) k^(n-1)
A(n)-k A(n-1)= K(1) h^(n-1)

kA(n)-hk A(n-1)= H(1) k^n
hA(n)-hk A(n-1)= K(1) h^n

(k-h)A(n)= H(1) k^n- K(1) h^n
A(n)= D k^n+ E h^n
Where D= H(1)/ (k-h)= (A(1)- h A(0))/(k-h)
E=- K(1)/(k-h)= K(1)/(h-k)= (A(1)- k A(0))/(h-k)

A(n) = B A(n-1) + C A(n-2)
A(n) - B A(n-1) - C A(n-2) = 0

The characteristic equation is x² - Bx - C = 0
Since h and k are the roots of the characteristic equation, if h≠k, A(n) is in the form:
A(n) = p*h^n + q*k^n, where p, q are two constants to be determined.

Put n = 0,
A(0) = p + q --------------(1)

Put n = 1,
A(1) = p*h + q*k ------------(2)

(2) - (1)*h:
A(1) - h A(0) = (k-h)*q
q = [A(1) - h A(0)] / (k-h)

Substituting into (1),
p = A(0) - q
= A(0) - [A(1) - h A(0)] / (k-h)
= [k A(0) - A(1)] / (k-h)

Therefore, A(n) = [k A(0) - A(1)] / (k-h) * h^n + [A(1) - h A(0)] / (k-h) * k^n


However, if h = k,
A(n) = (p + nq) h^n

Put n = 0,
A(0) = p

Put n = 1
A(1) = (p+q) h
A(1) = (A(0) + q) h
q = A(1)/h - A(0)

Therefore, A(n) = [A(0) + n(A(1)/h - A(0))]h^n