Proving a Mathematical Induction Problem Involving Arithmetic Sequences
The purpose of this article is to demonstrate the application of mathematical induction to prove the correctness of a statement involving the sum of an arithmetic sequence. The statement we aim to prove is: the sum of the first n positive integral values of the sequence 1, 4, 7, ..., (3n-2) can be expressed as (frac{n(3n-1)}{2}). This involves a series of logical steps to ensure the formula holds for all positive integers n.Step-by-Step Proof Using Mathematical Induction
We start by verifying the base case, i.e., when (n 1), the sum is 1. We then proceed to assume the statement is true for some (n k) and prove it for (n k 1).Step 1: Base Case
For (n 1), the sequence is just 1. The sum is 1, and we check the formula: [frac{1(3 cdot 1 - 1)}{2} frac{1 cdot 2}{2} 1.] This confirms the base case.Step 2: Inductive Hypothesis
Assume the statement is true for (n k), i.e., [1 4 7 dots (3k - 2) frac{k(3k - 1)}{2}.]Step 3: Inductive Step
We need to prove that the statement is also true for (n k 1), i.e., [1 4 7 dots (3k - 2) (3k 1) frac{(k 1)(3(k 1) - 1)}{2}.] From the inductive hypothesis, we have: [1 4 7 dots (3k - 2) frac{k(3k - 1)}{2}.] Adding (3k 1) to both sides, we get: [1 4 7 dots (3k - 2) (3k 1) frac{k(3k - 1)}{2} 3k 1.] Simplifying the right side, we have: [frac{k(3k - 1)}{2} 3k 1 frac{k(3k - 1) 2(3k 1)}{2}.] Further simplification gives us: [frac{3k^2 - k 6k 2}{2} frac{3k^2 5k 2}{2}.] Factoring the numerator, we get: [frac{3k^2 3k 2k 2}{2} frac{3k(k 1) 2(k 1)}{2} frac{(k 1)(3k 2)}{2} frac{(k 1)(3(k 1) - 1)}{2}.] This confirms that the statement is also true for (n k 1) if it is true for (n k). By the principle of mathematical induction, the statement is true for all positive integers (n).