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```					Tutorial 5 – Mathematical Induction COMP SCI 1FC3
Authors: Dai Tri Man Lê & Aaron Simpson

1. Simple Induction
To review, mathematical induction is a method that allows one to prove that a statement is true for all natural numbers. Using induction means that by proving the first statement (in an infinite sequence of statements) is true, and then proving that the “next” statement after any given point is true, then all must be true. Example : Use the Principle of Mathematical Induction to prove: ∑nk=1 (2k + 3) = n(n + 4) (for all n ≥ 1). Solution. Assume P(n) : 5 + 7 + 9 + ··· + (2n + 3) = n(n + 4) Base Step: P(1) states that 2 · 1 + 3 = 1(1 + 4), which is true since both sides equal 5. Inductive Hypothesis: Suppose P(k) is true; that is 5 + 7 + 9 + ··· + (2k + 3) = k(k + 4) (*) Inductive Step: P(k) → P(k+1): We want to show P(k+1) : 5+7+9+ ··· +(2k+3)+(2k+5) = (k+1)(k+5). To do so, we add 2k + 3 to both sides of (*) and simplify: LHS = 5 + 7 + 9 + ··· + (2k + 3) + (2k + 5) = k(k + 4) + (2k + 5) = k2 + 6k + 5 = (k + 1)(k + 5) = RHS which proves P(k + 1). Thus, P(k) ! P(k + 1) is true. Therefore, by the Principle of Mathematical Induction, 5+7+9+···+(2n+3) = n(n+4) is true for all n ≥ 1. (End of Solution.). You can use Maple's ability to manipulate expressions symbolically to help to construct an inductive proof. Here is how an interactive proof of the formula above, by mathematical induction, can be carried out in Maple.

Induction in Maple : The general term of the sum is term := k -> 2*k+3; while the right hand side of the formula is formula :=n -> n * (n + 4); We can check the basis step for the induction; here the base case is that in which n = 1. term(1); formula(1); The results agree, so the basis step is established. For the inductive step, we suppose the formula to be valid for n = k. indhyp := formula(k); To sum k + 1 terms, we compute the value of LHS LHS:=indhyp + term(k+1); Finally, the formula for n = k + 1 is RHS:=formula(k+1); To check if the results agree, we simplify the difference of LHS and RHS symbolically using the simplify Maple command. simplify(LHS-RHS); So the inductive step is verified. The formula now follows by mathematical induction. Thus, you can see that, while Maple is not (yet) able to construct proofs entirely on its own, it is a very effective tool to use in an interactive proof construction or symbolic computation. We were able to accomplish this using only simple assignment statements which we were already familiar with. Exercises: Use the Principle of Mathematical Induction to prove: 1. ∑nk=1 k = n(n + 1)/2 (for all n ≥ 1) 2. ∑nk=1 k = n(n + 1)/2 (for all n ≥ 1)

2. Strong Induction
Simple induction involved taking a single “base” case, and proving the next case in the sequence on that single foundation. By contrast, strong induction involves proving the next case in the sequence based on every previous case, not just a particular one. More formally: Whenever P(m), P(m+1), p(m+2),...,P(n) are all true, then P(n+1) is also true. Therefore P(n) is true for all integers n ≥ k. It's important to remember this difference, as otherwise the methodology used is the same as for “weak” induction. Exercises: Solve the weak induction section's problems using the strong induction format.

3. Take home problems: Sequences and Sums
Define the following sequences in Maple: 1. The sequence {cn}, where cn = n! 2. The sequence {dn}, where dn = (1 – n)2 + (1 – n)3 3. The sequence {en}, where en = n + 3n + 3n3 4. The sequence {fn}, where fn = (n + 6)/n – 1 Find : c100 6. en – en-1 7. fn/ f0 8. cn(n+1)(n+2)
5.

Also, compute the closed-forms using Maple. Prove by induction that these are correct. Define the following sums in Maple: 9. ∑nk=1 2k + 3 10. ∑nk=1 k2 11. ∑nk=1 (1 – k)2 + (1 – k)3 12. ∑nk=1 k/(k+1) 13. ∑∞k=1 1/k(k+1) n-1 14. ∑ k=1 ak – ai+1

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