Question

Solve the following recurrence relations.

a. x(n)=x(n-1)+5 for n 1, x(1)=0
b. x(n)=3x(n-1) for n>1, x(n)=4

Answer 1

To solve the given recurrence relations, the first is a linear sequence with a constant difference, resulting in x(n) = 5*(n-1). The second is a geometric sequence which yields x(n) = 4*3^(n-1) for each n>1.

Step-by-step explanation:

To solve these recurrence relations, let's tackle each one individually.

1. For the sequence x(n) = x(n-1) + 5, with initial condition x(1) = 0, it is a linear homogeneous recurrence relation with constant coefficients. We can observe that the difference in consecutive terms is always 5, and given that the first term is 0, this suggests that the nth term can be obtained by multiplying the difference with (n-1) (since the first term x(1) corresponds to n=1). Hence, x(n) = 5*(n-1).

2. For the sequence x(n) = 3*x(n-1), with the boundary condition x(1) = 4, we are dealing with a geometric progression where each term is three times the previous term. The nth term can be written as x(n) = 4*3(n-1), since each step multiplies the previous term by 3 and the initial term is 4.

Examining the structure of a recurrence relation helps us find the general formula and solve the entire sequence.