Problem 9 - A Graphical Approach to College Algebra

Question

Use mathematical induction to prove each statement. Assume that \(n\) is a positive integer. $$3+3^{2}+3^{3}+\cdots+3^{n}=\frac{3\left(3^{n}-1\right)}{2}$$

Step-by-Step Solution

Answer

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

Step 1: Base Case

First, we prove the base case where \( n = 1 \). Substitute \( n = 1 \) into the left-hand side of the equation: \[3 = \frac{3(3^1 - 1)}{2}\] Calculating the right hand side, we have:\[\frac{3(3 - 1)}{2} = \frac{3 \times 2}{2} = 3\]Since the left-hand side equals the right-hand side, the base case holds.

Step 2: Inductive Hypothesis

Assume the statement is true for some positive integer \( k \). This means we assume:\[3 + 3^2 + 3^3 + \cdots + 3^k = \frac{3(3^k - 1)}{2}\]This assumption will be used to prove that the statement is also true for \( k+1 \).

Step 3: Inductive Step

Using the inductive hypothesis, prove the statement is true for \( n = k+1 \).Start with the left-hand side for \( n = k+1 \): \[3 + 3^2 + 3^3 + \cdots + 3^k + 3^{k+1}\]Using the inductive hypothesis, substitute the equivalent expression:\[\frac{3(3^k - 1)}{2} + 3^{k+1}\]Combine terms with a common denominator:\[\frac{3(3^k - 1)}{2} + \frac{2 \times 3^{k+1}}{2} = \frac{3 \cdot 3^k - 3 + 2 \cdot 3^{k+1}}{2}\]Factor out \( 3^k \):\[\frac{3^k (3 + 2 \cdot 3) - 3}{2} = \frac{3^k \, 3^2 - 3}{2} = \frac{3^{k+1} \times 3 - 3}{2}\]Simplify:\[\frac{3(3^{k+1} - 1)}{2}\]This is the right-hand side of the expression for \( n = k+1 \), confirming the statement is true for \( n = k+1 \).

Step 4: Conclusion

Since the base case is true and the inductive step shows that if the statement is true for \( n = k \), then it is true for \( n = k + 1 \), by mathematical induction, the statement holds for all positive integers \( n \).

Key Concepts

Base Case

In mathematical induction, the base case is the starting point of our proof. For the given problem, we begin with proving the base case where \( n = 1 \). This step is crucial because it demonstrates that the formula is valid for the initial value of the series.

When \( n = 1 \), the series on the left-hand side simplifies to just \( 3 \). We then substitute \( n = 1 \) into the right-hand side of the equation:

• Calculate: \( \frac{3(3^1 - 1)}{2} \).
• This simplifies to: \( \frac{3(2)}{2} = 3 \).

Since both sides match, the base case holds true. This validation ensures that the formula works starting from the smallest positive integer.

Inductive Hypothesis

The inductive hypothesis is a critical assumption in the process of mathematical induction. In this step, we assume the statement is true for some positive integer \( k \). This assumption forms the bridge to prove the statement for the next integer, \( k+1 \).

For our specific exercise, the inductive hypothesis assumes:

• \( 3 + 3^2 + 3^3 + \cdots + 3^k = \frac{3(3^k - 1)}{2} \)

This assumption does not yet prove anything for \( k+1 \), but it sets the stage for the next crucial part, the inductive step. Think of it as laying down the hypothesis that allows us to climb one more step in the proof.

Inductive Step

In the inductive step, we use the inductive hypothesis to prove the formula for \( n = k+1 \). This step is essential to extend the validity of the formula from \( k \) to \( k+1 \). It is the heart of mathematical induction because it demonstrates the formula's sustainability for all positive integers.

Starting with the expression for \( n = k+1 \):

• The series becomes \( 3 + 3^2 + 3^3 + \cdots + 3^k + 3^{k+1} \).
• Applying the inductive hypothesis, replace the initial segment with \( \frac{3(3^k - 1)}{2} \).
• Add \( 3^{k+1} \) and simplify by getting a common denominator.
• Rearrange and simplify to show: \( \frac{3(3^{k+1} - 1)}{2} \).

Since we've proven it true for \( k+1 \), this verifies the statement holds for every subsequent positive integer.

Positive Integer

At the foundation of mathematical induction is the concept of positive integers. In this context, a positive integer is any whole number greater than zero. Induction typically applies to things we count, like 1, 2, 3, and so on.

The premise of induction assumes a statement is true for an initial positive integer and proves that if it holds for one integer, it holds for the next.
This leap from one integer to the next is the driving force that lets us conclude the statement is valid for all positive integers. Ensuring a "chain reaction," this method confirms that the formula holds no matter which positive integer you start with.