Section 3.3

• 2. a) f(1) = -6, f(2) = 12, f(3) = -24, f(4) = 48, f(5) = -96. b) f(1) = 16, f(2) = 55, f(3) = 172, f(4) = 523, f(5) = 1576. c) f(1) = 1, f(2) = -3, f(3) = 13, f(4) = 141, f(5) = 19,597. d) f(n) = 3.
• 4. Many answers are possible. a) f(2) = 2, f(3) = 5, f(4) = -1, f(5) = 0. b) f(n) = 1. c) f(2) = 2, f(3) = 5, f(4) = 33, f(5) = 1214. d) f(n) = 1.
• 6. a) a(n+1) = a(n) + 4, a(1) = 2. b) a(1) = 0, a(2) = 2, a(n) = a(n-2). c) a(1) = 2, a(n) = a(n-1) + 2n., d) a(1) = 1, a(n) = a(n-1) + 2n - 1.
• 8. Sm(n+1) = Sm(n) + 1, Sm(0) = m.
• 10. Base case for n = 1 is clear. Prove the inductive hypothesis: given the truth for n, prove it is true for n+1.
• 14. Base case for n=1 is clear. Prove the inductive hypothesis by replacing terms by the definition of the Fibonacci sequence where it helps simplify.
• 16. A^1 is the base case and is true. For the inductive step, A^(n+1) = A A^n. Prove this by multiplying the matrices.
• 22. a) 1 is in S and if n is in S, then n+2 is in S. b) 3 is in S and if n is in S, then 3n is in S. c) all integers are in S and if p(x) is in S, then so is x p(x)+n for integer n. This uses Horner's method. You can also build this by powers of x.
• 28. Base case (w1 lambda)R = lambdaR w1R can be proved. For the inductive step, let w2 = w3x, where w3 is of length on less than w2. (w1w2)R = (w1w3x)R = x(w1w3)R = (w3x)R w1R = w2R w1R.
• 30. Base cases, lambda and x are both palindromes. If alpha is a palindrome, then so is x alpha x.
• 36. a) 0, b) 2, c) 2, d) 4.
• 38. Base case for n=1 is true. If A(1,n) = 2^n, then A(1,n+1) = A(0,A(1,n)) = A(0,2^n) = 2^(n+1).
• 44. Let P(n) be the statement "F is well-defined at n." P(0) is true, because F(0) is specified. Assume P(n) to be true. Then F is well-defined at n+1, since F(n+1) is given in terms of F(n). Therefore by mathematical induction, P(n) is true for all n, i.e., F is well-defined as a function on the set of all nonnegative integers.