Problem 15 - A Graphical Approach to College Algebra

Question

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

Step-by-Step Solution

Answer

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

Step 1: Base Case

First, check the base case where \( n = 1 \). Substitute \( n = 1 \) into the left side of the equation: \( \frac{4}{5} \). Now substitute \( n = 1 \) into the right side: \( 1 - \frac{1}{5^{1}} = 1 - \frac{1}{5} = \frac{4}{5} \). The base case holds true as both sides equal \( \frac{4}{5} \).

Step 2: Inductive Hypothesis

Assume the statement is true for \( n = k \), i.e., assume \( \frac{4}{5} + \frac{4}{5^2} + \cdots + \frac{4}{5^k} = 1 - \frac{1}{5^k} \). This is the inductive hypothesis.

Step 3: Inductive Step

Prove the statement for \( n = k+1 \). Start with the left side for \( n = k+1 \): \[ \frac{4}{5} + \frac{4}{5^2} + \cdots + \frac{4}{5^k} + \frac{4}{5^{k+1}}.\] Using the inductive hypothesis, this simplifies to \[ \left( 1 - \frac{1}{5^k} \right) + \frac{4}{5^{k+1}}. \] Simplify further: \[ 1 - \frac{1}{5^k} + \frac{4}{5^{k+1}} = 1 - \frac{1}{5^k} + \frac{4}{5 \times 5^k} \].

Step 4: Simplify Rational Expression

Combine terms: \[ 1 - \frac{1}{5^k} + \frac{4}{5^{k+1}} = 1 - \frac{1}{5^k} + \frac{4}{5^{k+1}} = 1 - \frac{5}{5^{k+1}} + \frac{4}{5^{k+1}} = 1 - \frac{1}{5^{k+1}}.\] Thus, the right side matches the pattern, proving the statement for \( n = k+1 \).

Step 5: Conclusion

Since the base case holds and assuming the statement for \( n = k \) implies it holds for \( n = k+1 \), by mathematical induction, the equation \( \frac{4}{5} + \frac{4}{5^2} + \cdots + \frac{4}{5^n} = 1 - \frac{1}{5^n} \) is true for all positive integers \( n \).

Key Concepts

Base Case

Mathematical induction begins with checking the simplest form of the problem. This is known as the base case. For our exercise, we first test when \( n = 1 \) to ensure the pattern holds from the start. In mathematical induction, verifying this step provides a foundation for further steps.

For \( n = 1 \), substitute into both sides of our equation:

• Left side: \( \frac{4}{5} \)
• Right side: \( 1 - \frac{1}{5} = \frac{4}{5} \)

The left and right sides match perfectly. This means our base case is valid, providing the jump-off point for induction.

Inductive Hypothesis

The inductive hypothesis is like making an educated guess, though with more certainty. You assume what you want to prove is true for an arbitrary positive integer, call it \( k \).

In this process, assume:
\[\frac{4}{5} + \frac{4}{5^2} + \cdots + \frac{4}{5^k} = 1 - \frac{1}{5^k}\]This step is crucial because it allows you to use this assumption to show the next step in your logic. By assuming it's true for \( n = k \), we set the stage to prove it for \( n = k + 1 \).

Inductive Step

Now, we shift from assumption to execution. In the inductive step, prove the statement for \( n = k+1 \) using the inductive hypothesis.

• Start with: \( \frac{4}{5} + \frac{4}{5^2} + \cdots + \frac{4}{5^k} + \frac{4}{5^{k+1}} \)
• Use the hypothesis: \( \left( 1 - \frac{1}{5^k} \right) + \frac{4}{5^{k+1}} \)
• Simplify: \( 1 - \frac{1}{5^k} + \frac{4}{5^{k+1}} = 1 - \frac{1}{5^{k+1}} \)

This mathematical simplification shows the statement holds for \( n = k+1 \). Hence, the pattern continues from \( k \) to \( k+1 \), inching closer to a general proof.

Geometric Series

A geometric series is a sequence of numbers where each term after the first is found by multiplying the previous one by a constant. In our case, this constant is \( \frac{1}{5} \). The series from the problem can be represented as:

• \( \frac{4}{5} + \frac{4}{5^2} + \frac{4}{5^3} + \cdots + \frac{4}{5^n} \)

This series has a common ratio, \( r = \frac{1}{5} \), between consecutive terms.

Understanding this allows us to sum the series using the formula for the sum of the first \( n \) terms of a geometric series:\[ S_n = a \frac{1-r^n}{1-r} \]where \( a \) is the first term in the sequence, and \( r \) is the common ratio. The equation proves invaluable in predicting the series' behavior and confirming our proved expression.