Question

Prove each proposition by using induction. You may find it helpful to write out the first few terms, as is done in part a.

a) For all n in N, 2 + 5 + 8 + ... + (3n-1) = Σ(i=1 to n) (3i-1) = (3n+1)

Answer 1

To prove the proposition using induction, we show that it holds for the base case and then assume it holds for n=k, and prove it holds for n=k+1. By applying this method, we can prove that the given formula holds for all values of n in the set of natural numbers.

Step-by-step explanation:

To prove the proposition using induction, we need to show that the formula holds for the base case (n = 1) and assume it holds for n=k, and then prove it holds for n=k+1.

Base case (n = 1): The left-hand side (LHS) is 2, and the right-hand side (RHS) is 3(1)+1=4. Therefore, the formula holds for n=1.

Inductive step: Assume the formula holds for n=k, i.e., 2 + 5 + 8 + ... + (3k-1) = (3k+1). Now we need to prove it holds for n=k+1. Adding (3k+2) to both LHS and RHS, we have: 2 + 5 + 8 + ... + (3k-1) + (3k+2) = (3k+1) + (3k+2) = 3(k+1)+1 = (3(k+1)+1), which matches the RHS. Therefore, the formula holds for n=k+1.

By the principle of mathematical induction, the formula holds for all values of n in the set of natural numbers (N).