Problem 13 - Precalculus with Limits

Question

In Exercises 11 - 24, use mathematical induction to prove the formula for every positive integer \( n \). \( 2 + 7 + 12 + 17 + \cdots + \left(5n - 3\right) = \dfrac{n}{2}\left(5n - 1\right) \)

Step-by-Step Solution

Answer

Following the principle of mathematical induction, it can be proved that for each positive integer \( n \), the sum of a series that increases by 5 with each term, starting at 2, returns \( \dfrac{n}{2}\left(5n -1\right) \). This is shown by establishing a base case and proving that if the formula holds for an arbitrary integer \( k \), it also holds for the next integer \( k+1 \).

Step 1: Base Case

First, check the base case: when \( n = 1 \), the formula simplifies to \( 2 \). Checking with the original equation, we find that when \( n = 1 \), it results in \( 2 \) as well, so the base case holds true.

Step 2: Inductive Hypothesis

Assume that the given formula is valid for some arbitrary positive integer \( k \), that is \( 2 + 7 + 12 + 17 + \cdots + \left(5k - 3\right) = \dfrac{k}{2}\left(5k - 1\right) \). This is our inductive hypothesis.

Step 3: Inductive Step

Show that the formula is valid for \( n = k + 1 \). First, add \( 5(k + 1) - 3 \) to both sides of the formula. On the left side, it can be rewritten as the sum \( 2 + 7 + 12 + 17 + \cdots + \left(5k - 3\right) + \left(5(k+1) - 3\right) \). On the right side, it will become \( \dfrac{k}{2}\left(5k -1\right) + 5(k+1) - 3 \). Simplify the right side and show that it equals to \( \dfrac{(k+1)}{2}\left(5(k+1) -1\right) \). If it does, then the formula holds for \( k+1 \) and the principle of mathematical induction is established.

Key Concepts

Proof by Induction

Imagine you're climbing a stairway. You know that you can reach the first step. Then, you realize that if you're standing on a particular step, you can reach the next one. This logic guarantees that you can climb up as high as you want, step by step. That's the essence of proof by induction, an essential reasoning technique in mathematics that confirms an infinite sequence of statements.

Induction consists of two main stages:

• Base Case: Proving the statement is true for the initial case, usually when = 1.
• Inductive Step: Assuming the statement is true for an arbitrary case k, and then proving it's also true for the next case, k+1.

This method becomes a powerful tool to establish truth across an infinite set of mathematical statements, such as formulas for sums or sequences - exactly like our arithmetic sequence in the given exercise.

Inductive Hypothesis

The inductive hypothesis is a strategic pause in our journey of induction, where we assume, without proving, that the given statement holds for an arbitrary step, say the kth step. It's like saying, 'Okay, let's suppose I'm on the kth step of my stairway, and everything is fine here.'

In our arithmetic sequence problem, the hypothesis is assuming the sum formula is true for k. Why do this? Because it sets up a logical domino effect. If we can prove that this assumption leads correctly to the k+1 case, then by the principle of induction, the formula must be true for all subsequent steps - sealing the deal for an infinitely long sequence.

Arithmetic Sequence

An arithmetic sequence is a list of numbers with a common difference between consecutive terms. Common everyday examples include the sequence of odd or even numbers. In our exercise, the numbers 2, 7, 12, etc., form an arithmetic sequence with a common difference of 5.

This sequence can be generalized as a, a+d, a+2d, ..., where a is the first term and d is the common difference. The formula for the nth term of such a sequence is a + (n-1)d, and the sum of the first n terms can be expressed as (n/2)*(first term + nth term). Remember, the familiarity with these concepts allows students to feel more comfortable with mathematical induction, as they provide the fundamental structures that induction often seeks to prove.