Problem 1 - College Algebra with Modeling and Visualization

Question

Use mathematical induction to prove the statement. Assume that \(n\) is a positive integer. $$ 3+6+9+\dots+3 n=\frac{3 n(n+1)}{2} $$

Step-by-Step Solution

Answer

The given formula is true for all positive integers by induction.

Step 1: Base Case

To prove the statement by induction, first verify the base case when \(n=1\). The left-hand side of the statement is simply \(3 cdot 1 = 3\). The right-hand side is \(\frac{3 \cdot 1(1+1)}{2} = 3\). Since both sides are equal, the base case holds.

Step 2: Inductive Hypothesis

Assume that the statement is true for some positive integer \(k\). That is, assume \[3 + 6 + 9 + \dots + 3k = \frac{3k(k+1)}{2}.\]

Step 3: Inductive Step

We need to show that the statement is true for \(k+1\). The left-hand side for \(k+1\) becomes \[3 + 6 + 9 + \dots + 3k + 3(k+1).\]By the inductive hypothesis, we know that \[3 + 6 + 9 + \dots + 3k = \frac{3k(k+1)}{2}.\]Therefore, the equation becomes\[\frac{3k(k+1)}{2} + 3(k+1).\]

Step 4: Simplify Inductive Step

Combine and simplify the expression: \[\frac{3k(k+1)}{2} + \frac{6(k+1)}{2} = \frac{3k(k+1) + 6(k+1)}{2}.\]Factor \(3(k+1)\) from the numerator: \[\frac{3(k+1)(k+2)}{2}.\]Thus, the statement holds for \(k+1\):\[3 + 6 + 9 + \dots + 3(k+1) = \frac{3(k+1)((k+1)+1)}{2}.\]

Step 5: Conclusion

Since the base case holds and the induction step is true, by the principle of mathematical induction, the statement is proved for all positive integers \(n\).

Key Concepts

Algebraic Proofs

Algebraic proofs are a fascinating and crucial part of mathematics, particularly when you need to verify statements or formulas. Using algebraic proofs involves logical reasoning to show that a particular equation or statement is true in all cases, given the defined constraints. In our exercise, mathematical induction is used as a method to prove that a formula holds for all positive integers.

• First, establish a base case to showcase that the formula works for an initial value.
• Next, assume the formula works for some integer "k"; this is your inductive hypothesis.
• Finally, prove it also holds for "k+1" by deduction. If these steps succeed, you've proven the statement for all integers.

Mathematical induction might seem a bit abstract initially, but it builds on simple subtraction and addition steps. This reinforces learning and helps gain deeper insights into the sequences involved.

Sequences and Series

Sequences and series are foundational concepts in mathematics. A sequence is simply a list of numbers following a certain pattern, while a series is the sum of the terms in a sequence. In our specific problem, the sequence is an arithmetic progression: 3, 6, 9, ..., 3n.

• Arithmetic sequences have a constant difference between consecutive terms. For our sequence, the difference is 3.
• A series formed from a sequence essentially calculates the cumulative total of terms from that sequence.

Understanding sequences and series not only supports solving mathematical problems but also enhances comprehension of real-world patterns and calculations. Recognizing the type of sequence or series is crucial to successfully applying the right formulas and proving statements, especially in algebraic proofs.

Sum of Sequence

Finding the sum of a sequence is a common task in mathematics, especially when dealing with a series. For arithmetic sequences, we often use formulas to bring together terms efficiently. In our example, the sum of the sequence 3, 6, 9, ..., 3n is innovatively calculated using the formula \( \frac{3n(n+1)}{2} \).

• This formula is derived from general principles of arithmetic sequences, specifically focusing on their mean and scaling.
• By understanding the formula, we can quickly determine the total of the sequence without having to sum each term individually.

The elegance of this approach lies in its simplicity yet powerful method for dealing with large data sets, enhancing both efficiency and accuracy in calculations. By applying these formulas, mastering complex arithmetic series becomes much more approachable.