1) First hole is played by the golfer Assuming the golfer plays Kth hole. Since he played the kth hole he must also play the k+1th (given) =>Therefore the golfer will play all the holes 2) a) P(1) is the statement 1^3 = ((1(1 + 1)/2)^2 b) Both sides of eqn evaluate to 1. c) Induction hypothesis is P(k) for positive integer k the statement 1^3+ 2^3+· · ·+k^3 = (k(k+ 1)/2)^2 d) Lets assume P(k) holds, we have to show that P(k + 1) is true, we have to derive equation 1^3+2^3+· · ·+k^3+(k+1)^3 = ((k + 1)(k + 2)/2)^2 e) Add (k + 1) ^3 to LHS and RHS of equation in induction hypothesis. Showing that LHS in d) is equal to (k(k + 1)/2) ^2 + (k + 1) ^3. We find this equals ((k + 1)(k + 2)/2)^2. Therefore LHS = RHS f) We did the basis step and the inductive step. Principle of mathematical induction now proves that P(n) is true for all positive integers n. 3) 4) Base Case(s): Notice that we can form 18 cents of postage using 6 3-cent stamps. We can form 19 cents of postage using 1 10-cent stamp and 3 3-cent stamps. We can form 20 cents of postage using2 10-cent stamps. Induction Step: Assume we form any postage i with 18 ≤ i ≤ n, with n ≥ 20.Consider forming n + 1 cents of postage. By strong induction hypothesis, since n ≥ 20, n − 2 ≥ 18, we then have n − 2 cents of postage. Adding single three cent stamp then gives n + 1 cents of postage. Hence proved 5) The assumption in the inductive step is that there always is minimum 1 3-cent stamp or minimum 2 4-cent stamps. Proof is also flawed since assumption isn’t guarantee in proposition and the inductive step fails to prove from four cents to five cents, as basis 4-cent case has no two 4-cent stamps and also no 3-cent stamps.