solve the recurrence an=-3a n-1 +10a n-2, n≥2 given a 0=1,a 1=4

Question

solution of this question.

Answer 1

solve the recurrence an=-3a n-1 +10a n-2, n≥2 given a 0=1,a 1=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)