Problem 7 - A Graphical Approach to Precalculus with Limits

Question

Use mathematical induction to prove each statement. Assume that \(n\) is a positive integer. $$5+10+15+\dots+5 n=\frac{5 n(n+1)}{2}$$

Step-by-Step Solution

Answer

The formula is true for all positive integers \( n \) by induction.

Step 1: Base Case Verification

First, verify the theorem for the smallest value of the positive integer, which is typically 1. For \( n = 1 \), the left-hand side of the equation is simply the first term, 5. Observe the right-hand side: \( \frac{5 \cdot 1 \cdot (1 + 1)}{2} = \frac{5 \cdot 2}{2} = 5 \). Hence, the base case holds true.

Step 2: Inductive Hypothesis Assumption

Assume that the formula holds for some arbitrary positive integer \( k \). That is, assume \( 5 + 10 + 15 + \dots + 5k = \frac{5k(k+1)}{2} \). This is our inductive hypothesis.

Step 3: Inductive Step Proof

We need to prove that if the statement is true for \( n = k \), it must also be true for \( n = k + 1 \). Consider the equation: \[ 5 + 10 + 15 + \dots + 5k + 5(k + 1) = \frac{5(k+1)(k+2)}{2} \].Using the inductive hypothesis, substitute:\[ \frac{5k(k+1)}{2} + 5(k+1) = \frac{5(k+1)(k+2)}{2} \].Factor out \( 5(k+1) \):\[ \frac{5(k+1)(k+2)}{2} = \frac{5(k+1)(k+2)}{2} \].This deduces equality, confirming the statement holds for \( n = k + 1 \).

Step 4: Conclusion of Induction

Having verified the base case and the inductive step, we conclude by the principle of mathematical induction that the statement is true for all positive integers \( n \).

Key Concepts

Series

A series is a summation of terms of a sequence, typically following a specific rule or pattern. In mathematics, a series can be infinite or finite. In this context, we are working with a finite series. Our sequence consists of multiples of five: 5, 10, 15, ..., up to a certain point, specified by the term \(5n\).
This series is arithmetic, where each term increases by a constant value. Recognizing the arithmetic nature helps us apply mathematical formulas for finding sums easily. Our goal is to prove the formula for the sum of this series using induction.
Understanding series and their properties allows for simplifications in calculations and provides a clearer path to solving advanced mathematical problems. By finding an efficient way to sum these numbers, we use less computational effort and can handle larger sequences with ease.

Sum of Multiples

Calculating the sum of multiples involves adding all numbers that are a multiple of a certain base, in this case, five.
The example involves multiples of five added sequentially: 5, 10, 15, ..., up to a term that is determined by a positive integer \(n\).
To express this sum mathematically, we use the formula \(5 + 10 + 15 + \, \ldots \, + 5n = \frac{5n(n+1)}{2}\). This formula simplifies the process of adding a series of multiples by converting a potentially long series of addition operations into a concise algebraic expression.
The approach reduces computational complexity, as calculating the result directly from the formula is much faster than manually summing each of the multiples. Understanding how to derive or prove such formulas is crucial in fields requiring large series calculations, such as economics and physics.

Inductive Hypothesis

The inductive hypothesis is a key part of proving statements using mathematical induction. It represents the step where we assume the statement is true for some arbitrary positive integer, usually denoted as \(k\).
In our problem, the inductive hypothesis assumes that the equation \(5 + 10 + 15 + \, \ldots \, + 5k = \frac{5k(k+1)}{2}\) holds true.
This assumption is strategic, as it sets up the foundation for proving the next step, which involves showing that if the statement is true for \(n = k\), it must also be true for \(n = k+1\).

• Make an assumption for an arbitrary point, \(k\).
• Consider this assumption as a basis for proving the next logical step.

Using this hypothesis simplifies the process by allowing you to build upon an assumed truth rather than starting from scratch at each step.

Base Case

In mathematical induction, the base case is the first step where you verify the truth of a statement for the initial value, typically \(n = 1\). This initial verification is crucial, as it ensures a solid start for your proof.
For our problem, we found that when \(n = 1\), both sides of the equation result in 5:

• Left-hand side: Just the number 5
• Right-hand side: \(\frac{5 \cdot 1 \cdot (1+1)}{2} = 5\)

This proves our base case is correct. Only when the base case holds can we proceed with the induction process. It serves as the foundational step upon which the validity of the rest of the proof is built. Skipping the base case results in an unverified induction proof, which could lead to incorrect conclusions about the formula's validity.