Question

10. (10 points) Computational problem solving: Proving correctness: Function g (n: nonnegative integer) if n si then return(n) else return(5*g(n-1) - 6*g(n-2)) Prove by induction that algorithm g is correct, if it is intended to compute the function 3"-2" for all n 20. Base Case Proof: Inductive Hypothesis: Inductive Step:

Answer 1

Base Case:

Let us assume that n = 1.

Hence, function g returns 1 since n <= 1.

Now, 3ⁿ - 2ⁿ = 3 - 2 = 1

Clearly, the equality is satisfied for base case n = 1.

Inductive Hypothesis:

Let us assume that for n <= k, the function g(k) returns 3^k - 2^k and g(k-1) = 3^k-1 - 2^k-1

Inductive Step:

Now, we would need to show that for n = k + 1, the function g(k + 1) returns 3^k+1 - 2^k+1.

Proof:

g(k + 1) = 5*g(k) - 6*g(k-1)

= 5 * (3^k - 2^k) - 6(3^k-1 - 2^k-1)

= 15*3^k-1 - 10*2^k-1 - 6*3^k-1 + 6*2^k-1

= 9*3^k-1 - 4*2^k-1

= 3^k+1 - 2^k+1 (proved)