Problem 12 - College Algebra

Question

In Exercises 11-24, use mathematical induction to prove that each statement is true for every positive integer \(n.\) $$3+4+5+\cdots+(n+2)=\frac{n(n+5)}{2}$$

Step-by-Step Solution

Answer

The formula \(3 + 4 + 5 +...+ (n + 2) = \frac{n(n + 5)}{2}\) is incorrect as it does not hold for every positive integer.

Step 1: Base Case

The base case for this problem is verifying the truth of the statement when \(n = 1\). On the left hand side of the statement, we get \(3 + 1 = 4\), and on the right hand side we have \(\frac{1(1 + 5)}{2} = 3\). So, the base case \(n = 1\) does not hold true.

Step 2: Assumption (Inductive Hypothesis)

The next step in the process is to assume that the formula is true for some arbitrary positive integer \(n = k\), which gives us \(3 + 4 + 5 + ... + (k + 2) = \frac{k(k + 5)}{2}\). This is the Inductive Hypothesis.

Step 3: Inductive Conclusion

Finally, we need to show that the statement holds for \(n = k + 1\), assuming it holds for \(n = k\). That would hence require us to show that \(3 + 4 + 5 + ... + (k + 2) + ((k+1) + 2) = \frac{(k + 1)((k + 1) + 5)}{2}\). However, from the incorrect base case, the formula does not hold for every positive integer.

Key Concepts

Base Case

In mathematical induction, you always start with the base case. This is the foundation of your proof, where you show the statement is true when the variable is at its starting value, typically 1. By checking this initial value, you can determine if your formula gives the expected, correct result right from the beginning.
The base case in our exercise checks the statement when \( n = 1 \). This means plugging 1 into both the left and right sides of the equation:

• Left side: The series starts from 3 and ends at \((1+2)\), giving \(3\).
• Right side: Calculating \(\frac{1(1+5)}{2} = 3\).

Unfortunately, although the example incorrectly states that the base case didn't match, both sides should indeed be 3 if calculated properly. The accurate evaluation sets a solid foundation for moving forward with the proof.

Inductive Hypothesis

An essential step in mathematical induction is the inductive hypothesis. This is where we assume the statement is true for an arbitrary positive integer. When we use this step, we call this integer \( k \). Assuming correctness for \( n = k \) helps us build a bridge to prove it for \( n = k+1 \).
In this context, the inductive hypothesis means accepting that:

• \(3 + 4 + 5 + ... + (k + 2) = \frac{k(k + 5)}{2}\)

This assumption simplifies the transition from one integer to the next. Although we're assuming here, this is a key part of how induction mirrors a domino effect, where the initial truth causes subsequent truths.

Inductive Step

The inductive step is where the magic happens in mathematical induction. Assuming the hypothesis is true, we now show it's also true for the next integer, \( n = k + 1 \).
For our exercise, the inductive step grows the sequence to include \((k+1)+2\), expanding our formula:

• Left side: \(3 + 4 + 5 + ... + (k + 2) + ((k+1) + 2)\)
• Right side: \(\frac{(k + 1)((k + 1) + 5)}{2}\)

If both sides balance, it solidifies the statement for all values up through \( k+1 \). Mathematically, this supports the validity of the formula for any positive integer, indicating that if it's true for \( n = k \), it holds boomingly for \( n = k + 1 \). This is the logical anchor that justifies the use of induction.

Positive Integers

Positive integers are at the heart of mathematical induction. When we work through induction, we're proving a statement for all numbers greater than zero. They sequentially follow one another (like 1, 2, 3, etc.), forming the backbone of induction.
In our problem, we're dealing with positive integers because these are numbers that are:

• Greater than zero.
• Whole and without fractions or decimals.
• Used continuously for step-by-step inferences in induction.

Understanding this is crucial because it contextualizes the proof. Starting with the smallest, simplest positive integer (often 1) allows mathematical induction to build a proof that applies to all such integers beyond that initial instance. By ensuring the statement is valid for this infinite set, induction becomes a powerful tool for proofs in mathematics.