Problem 8 - A Graphical Approach to Precalculus with Limits

Question

Use mathematical induction to prove each statement. Assume that \(n\) is a positive integer. $$4+7+10+\dots+(3 n+1)=\frac{n(3 n+5)}{2}$$

Step-by-Step Solution

Answer

The statement is proven for all positive integers using mathematical induction.

Step 1: Base Case

We first need to check if the statement holds for the smallest positive integer, which is usually 1. Substitute \( n = 1 \) into both sides of the equation. The left-hand side (LHS) calculates as: \( 3(1) + 1 = 4 \).The right-hand side (RHS) calculates as: \( \frac{1(3(1) + 5)}{2} = \frac{1 \times 8}{2} = 4 \).Since the LHS equals the RHS, the base case holds.

Step 2: Inductive Hypothesis

Assume that the statement holds true for some positive integer \( n = k \). That is, assume:\[4 + 7 + 10 + \dots + (3k + 1) = \frac{k(3k + 5)}{2}\]

Step 3: Inductive Step

Now, prove that the statement also holds for \( n = k + 1 \). We need to show:\[4 + 7 + 10 + \dots + (3k + 1) + (3(k+1) + 1) = \frac{(k+1)(3(k+1) + 5)}{2}\]Using the inductive hypothesis:\[4 + 7 + 10 + \dots + (3k + 1) = \frac{k(3k + 5)}{2}\]Add \( 3(k+1) + 1 \) on both sides:\[\frac{k(3k + 5)}{2} + (3k + 4)\]

Step 4: Simplify the Expression

Simplify the expression:\[\frac{k(3k + 5) + 2(3k + 4)}{2} = \frac{3k^2 + 5k + 6k + 8}{2} = \frac{3k^2 + 11k + 8}{2}\]Simplify the RHS for \( n = k + 1 \):\[\frac{(k+1)(3(k+1) + 5)}{2} = \frac{(k+1)(3k + 3 + 5)}{2} = \frac{(k+1)(3k + 8)}{2}\]

Step 5: Verify Equality

We must verify that:\[3k^2 + 11k + 8 = (k^2 + k)(3) + (k + 1)8\]Which simplifies to:\[3k^2 + 11k + 8\]Since both expressions for \( n = k+1 \) match, the induction step holds.

Step 6: Conclusion

Since both the base case is true, and assuming the statement for \( n = k \) implies it is true for \( n = k+1 \), by the principle of mathematical induction, the statement is true for all positive integers \( n \).

Key Concepts

Sequence Summation

Summing sequences is a fundamental concept in mathematics, especially when dealing with series that follow a regular pattern. In our problem, we have a sequence expressed as \(4, 7, 10, \ldots, (3n + 1)\).
This is an arithmetic sequence where each term increases by a constant difference of 3. Recognizing this pattern allows us to tackle problems systematically rather than recalculating every term.Here's how this works:

• The first term is 4, which corresponds to \(3 \times 1 + 1\).
• The second term is 7, calculated as \(3 \times 2 + 1\).
• This pattern continues, as each new term increases by 3.

Understanding this pattern is key to summing up to the \(n\)-th term efficiently. Ultimately, problems like these simplify when you identify the sequence's rule or pattern.

Base Case

In mathematical induction, verifying the base case is crucial as it forms the foundation for the entire proof. The base case is typically the smallest value of the positive integer \(n\) for which the statement needs confirmation.

For our specific problem, we set \(n = 1\) and check both sides of the equation:

• The left-hand side (LHS) becomes just the first term: \(3 \times 1 + 1 = 4\).
• The right-hand side (RHS) evaluates the expression: \(\frac{1 (3 \times 1 + 5)}{2} = 4\).

Since both sides match, the base case holds true. Verifying this alignment accurately solidifies the statement's reliability for higher values of \(n\). It sets the premise for the inductive step to extend upon.

Inductive Hypothesis

In mathematical induction, the inductive hypothesis is a crucial step where you assume the statement is true for some arbitrary positive integer \(n = k\).

For our context, we assume:

• \(4 + 7 + 10 + \cdots + (3k + 1) = \frac{k(3k + 5)}{2}\)

This assumption is important because it helps us bridge to the next step. By assuming the hypothesis is true for \(n = k\), we lay the groundwork to eventually prove it true for \(n = k+1\).
By having this assumption in place, your focus shifts to showing that if it holds true for \(k\), it should logically hold for the following integer.

Inductive Step

In this phase of mathematical induction, we aim to demonstrate that if a statement is true for \(n = k\), it should also be true for \(n = k+1\).

Here's a breakdown of how this works:

• Start with the assumption from the inductive hypothesis: \(4 + 7 + 10 + \cdots + (3k + 1) = \frac{k(3k + 5)}{2}\).
• Add the next term, \((3(k+1) + 1)\), to both sides. This step is crucial for moving from \(n=k\) to \(n=k+1\).
• Simplify the combined expression: \(\frac{k(3k+5) + 2(3k+4)}{2}\) and manipulate it to match the form \(\frac{(k+1)(3k+8)}{2}\).

By verifying both the derived expression and the targeted expression coincide, you complete the proof. This validates that the original statement holds true for the subsequent integer, thus establishing the pattern continues infinitely for all \(n\).