solve the recurrence an=-3an-1 +10an-2, n≥2 given a0=1,a1=4
The characteristic equation is,
s^2 + 3s – 10 = 0
(s + 5)(s – 2) = 0
s = -5, s = 2
The recurrence relation then is: an = A.(-5)^n + B.(2)^n
Initial conditions
n = 0: a0 = 1 = A.1 + B.1 à 1 = A + B
n = 1: a1 = 4 = A.(-5) + B.(2) à 4 = -5A + 2B
Subtracting twice the 1st equation from the 2nd equation gives,
2 = -7A
A = -2/7, and B = 9/7
The recurrence relation now becomes: an = (1/7)(9.(2)^n – 2.(-5)^n)