Problem 16 - A Graphical Approach to College Algebra

Question

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

Step-by-Step Solution

Answer

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

Step 1: Base Case

First, we prove the base case where \( n = 1 \). The left side of the equation becomes \( \frac{1}{2} \). The right side evaluates as \( 1 - \frac{1}{2} = \frac{1}{2} \). Since both sides are equal, the base case holds.

Step 2: Induction Hypothesis

Assume the statement is true for some positive integer \( k \), meaning \( \frac{1}{2} + \frac{1}{2^2} + \cdots + \frac{1}{2^k} = 1 - \frac{1}{2^k} \). This is the induction hypothesis.

Step 3: Inductive Step

We need to prove that if the statement is true for \( n = k \), then it is also true for \( n = k + 1 \). Consider the left-hand side for \( n = k + 1 \):\[ \frac{1}{2} + \frac{1}{2^2} + \cdots + \frac{1}{2^k} + \frac{1}{2^{k+1}}. \] By the induction hypothesis, this equals: \[ \left(1 - \frac{1}{2^k}\right) + \frac{1}{2^{k+1}}. \] Combine terms on the right:\[ 1 - \frac{1}{2^k} + \frac{1}{2^{k+1}} = 1 - \frac{1}{2^k} + \frac{1}{2 \cdot 2^k} = 1 - \frac{1}{2^k} + \frac{1}{2^{k+1}}. \]The common denominator for these fractions is \( 2^{k+1} \), so:\[ 1 - \frac{2}{2^{k+1}} + \frac{1}{2^{k+1}} = 1 - \frac{1}{2^{k+1}}. \]Thus, the statement holds for \( n = k + 1 \).

Step 4: Conclusion

Since the base case holds and the inductive step has been proven, by mathematical induction, the statement \( \frac{1}{2} + \frac{1}{2^2} + \cdots + \frac{1}{2^n} = 1 - \frac{1}{2^n} \) holds for all positive integers \( n \).

Key Concepts

Proof by Induction

Proof by induction is a powerful method in mathematics used to prove statements, especially those involving a natural number, typically denoted as \( n \). It consists of two major steps:
the base case and the inductive step.

This method is analogous to dominoes falling in line. You assert that the first domino falls (the base case), and then show that if one domino falls, the next does too (the inductive step). If both these conditions are satisfied, then all dominoes will eventually fall, meaning the statement is true for all natural numbers.

Induction is crucial because it extends the validity of a statement from a specific case to an infinite sequence of cases, making it an invaluable tool in proofs that involve series or sequences.

Geometric Series

A geometric series is a sequence where each term after the first is found by multiplying the previous term by a fixed, non-zero number called the common ratio. In the exercise provided, we have a geometric series with a common ratio of \( \frac{1}{2} \).

The series in the problem can be expressed as:

• First term: \( \frac{1}{2} \)
• Second term: \( \frac{1}{4} \)
• Third term: \( \frac{1}{8} \)
• And so on...

The general formula for the sum \( S_n \) of the first \( n \) terms of a geometric series is:
\[ S_n = a \frac{1 - r^n}{1 - r} \]
where \( a \) is the first term and \( r \) is the common ratio. In the given problem, we want to show that this formula matches the expression \( 1 - \frac{1}{2^n} \).

Base Case

The base case is the first step in proof by induction. It acts like the anchor of the proof, confirming that the statement holds true for the initial value of the sequence or series.

In the problem at hand, the base case involves evaluating the statement for \( n = 1 \). We calculate the left-hand side and right-hand side separately:

• The sum is \( \frac{1}{2} \) for \( n = 1 \).
• The expression \( 1 - \frac{1}{2} = \frac{1}{2} \).

This confirms that the statement is true for \( n = 1 \). Verifying the base case ensures that the process of induction has a correct starting point.

Inductive Step

The inductive step is a critical part of proving statements using induction. This step is where we show that if the statement holds for a certain case \( n = k \), then it must also hold for the next case \( n = k+1 \).

Here's how this is done in our exercise:

• Assume the statement is true for \( n = k \), meaning the series sums to \( 1 - \frac{1}{2^k} \).
• Next, show that including one more term yields \( 1 - \frac{1}{2^{k+1}} \).
• This involves adding \( \frac{1}{2^{k+1}} \) to the assumed sum, and manipulating the expression to reach the desired result.

This shows the relationship between consecutive terms, proving that if one part of the chain is true, the next is as well. Successfully completing the inductive step demonstrates that the entire series holds true for all \( n \).