Solve the following recurrence relations.a. x(n) = x(n − 1) + 5 for n > 1, x(1) = 0b. x(n) = 3x(n − 1) for n > 1, x(1) = 4c. x(n) = x(n − 1) + n for n > 0, x(0) = 0d. x(n) = x(n/2) + n for n > 1, x(1) = 1 (solve for n = 2k)e. x(n) = x(n/3) + 1 for n > 1, x(1) = 1 (solve for n = 3k)

Sure, let's solve these recurrence relations one by one:

a. x(n) = x(n − 1) + 5 for n > 1, x(1) = 0

This is a linear homogeneous recurrence relation with constant coefficients. The solution is given by the sum of the particular solution and the homogeneous solution. In this case, the homogeneous solution is 0 (since x(1) = 0), and the particular solution is a linear function. By substituting n = 1, 2, 3, ..., we can see that x(n) = 5*(n-1).

b. x(n) = 3x(n − 1) for n > 1, x(1) = 4

This is also a linear homogeneous recurrence relation with constant coefficients. The solution is given by x(n) = C*3^n, where C is a constant. By substituting x(1) = 4, we find that C = 4/3. So, x(n) = (4/3)3^n = 43^(n-1).

c. x(n) = x(n − 1) + n for n > 0, x(0) = 0

This is a linear non-homogeneous recurrence relation. The homogeneous solution is 0 (since x(0) = 0), and the particular solution can be found by guessing a form based on the non-homogeneous term. In this case, we guess that the particular solution is of the form an^2 + bn + c. By substituting and comparing coefficients, we find that a = 1/2, b = 1/2, and c = 0. So, x(n) = (1/2)n^2 + (1/2)n.

d. x(n) = x(n/2) + n for n > 1, x(1) = 1 (solve for n = 2k)

This is a non-linear homogeneous recurrence relation. By substituting n = 2^k, we get x(2^k) = x(2^(k-1)) + 2^k. This can be solved by defining a new sequence y(k) = x(2^k)/2^k, which satisfies the linear recurrence relation y(k) = y(k-1) + 1. The solution to this is y(k) = k + C, where C is a constant. By substituting y(0) = x(1) = 1, we find that C = 1. So, y(k) = k + 1, and x(n) = n*(log2(n) + 1).

e. x(n) = x(n/3) + 1 for n > 1, x(1) = 1 (solve for n = 3k)

This is also a non-linear homogeneous recurrence relation. By substituting n = 3^k, we get x(3^k) = x(3^(k-1)) + 1. This can be solved by defining a new sequence y(k) = x(3^k), which satisfies the linear recurrence relation y(k) = y(k-1) + 1. The solution to this is y(k) = k + C, where C is a constant. By substituting y(0) = x(1) = 1, we find that C = 1. So, y(k) = k + 1, and x(n) = log3(n) + 1.