Question

Let q ∈ R. Suppose that q > 0. Prove that 1+nq ≤ (1+q)n for all n ∈ N. We prove the result by induction on n. First suppose that n=1 So, 1+q = (1+q)1. This tells us the base case is true for n=1. Now we have to prove the inductive step. Lets assume this is the case for n +1 1 + (n+1)q = 1 + nq + q ≤ (1 +q)n +q (1+q)n +q – (1+q)n+1 = (1+q)n (1-1-q) +q = -q (1+q)n +q < 0 (1 +q)n +q < (1+q)n+1 1+ (n+1)q ≤ (1+q)n+1 Thus we have concluded that it is true for n+1 Proving true for all n ∈ N. This is what I have, but can someone help me with commentary and walk me through with explanations so I understand?

Answer 1

We use mathematical induction to prove that 1+nq ≤ (1+q)^n for all natural numbers n and positive real number q. The base case is verified for n=1, and the inductive step is proven using the inductive hypothesis and binomial theorem.

Step-by-step explanation:

We are tasked with using mathematical induction to prove that for all n in the natural numbers, 1+nq ≤ (1+q)^n given that q is a positive real number. The base case for n=1 is straightforward, as 1+q = (1+q)^1, which confirms our base case. For the inductive step, we assume that the inequality holds for some natural number n; that is, we assume 1+nq ≤ (1+q)^n is true. To prove it for n+1, we evaluate 1+(n+1)q = 1 + nq + q and use our inductive hypothesis to show 1 + nq + q ≤ (1+q)^n +q.

Next, we aim to prove that (1+q)^n + q ≤ (1+q)^{n+1}. The inequality stems from the expansion of the binomial theorem and the fact that q, and therefore 1+q, is positive. This allows us to conclude 1+(n+1)q ≤ (1+q)^{n+1}, completing the inductive step. Consequently, we've shown the inequality holds for all natural numbers n. The binomial theorem and properties of inequalities are crucial to the argument.