Problem 19 - College Algebra

Question

In Exercises 11-24, use mathematical induction to prove that each statement is true for every positive integer \(n.\) $$2+4+8+\dots+2^{n}=2^{n+1}-2$$

Step-by-Step Solution

Answer

The statement \(2+4+8+\dots+2^{n}=2^{n+1}-2\) is proven to be true for all positive integers by using the principle of mathematical induction. This is done by showing it is true for \(n=1\) (base case), assuming it holds for some positive integer \(n=k\), and proving it holds for \(n=k+1\).

Step 1: Identify and prove the base case

The base case normally starts at \( n = 1 \). So, we substitute \( n = 1 \) into both sides of the equation \(2+4+8+\dots+2^{n}=2^{n+1}-2\) and ensure equality holds. This only results in the term \(2^1\) on the left side, and \(2^{1+1}-2\) on the right side. Therefore we have 2(1) = 2(2) - 2 which simplifies to \(2 = 2\), confirming that our formula holds for the base case.

Step 2: Express the induction hypothesis

The induction hypothesis is an assumption that the equation \(2+4+8+\dots+2^{n}=2^{n+1}-2\) is true for some positive integer \(k\). So, we have \(2+4+8+\dots+2^{k}=2^{k+1}-2\) based on the assumption.

Step 3: Carry out the induction step and prove for \(n = k + 1\)

The induction step is the key part of mathematical induction, where we add the new term for the next value of \(n\), \(k + 1\), to both sides of the equation, and demonstrate that the equation still holds. We add \(2^{k+1}\) to both sides of the equation, resulting in \(2 + 4 + 8 + \dots + 2^k + 2^{k+1} = 2^{k+1} - 2 + 2^{k+1}\). After simplifying this equation, the right-hand side matches the left-hand side, demonstrating the validity of the assumed formula for \(n = k + 1\).

Step 4: Conclusion

Since we have shown that the statement holds for \(n = 1\) (base case), and we have assumed that it holds for some \(n = k\) and have proven it for \(n = k + 1\), we conclude by the principle of mathematical induction that the statement \(2+4+8+\dots+2^{n}=2^{n+1}-2\) is true for all positive integers.

Key Concepts

Base Case

Mathematical induction kicks off by establishing a starting point, known as the base case. In this scenario, the smallest positive integer, typically referred to as \( n = 1 \), is used. Here’s a breakdown on why this step is crucial:

• We substitute \( n = 1 \) into our given equation to check its validity for the very first number in our sequence. In our example, the left side yields \( 2 \) since it only considers the term \( 2^1 \).
• The right side, substituting the same \( n=1 \), becomes \( 2^{1+1} - 2 = 2^2 - 2 = 4 - 2 = 2 \).

Both sides equate to \( 2 \), confirming that the base case holds true. This step sets a solid foundation for the ensuing steps by confirming that our formula is accurate for the initial step of the induction process.

Induction Hypothesis

The induction hypothesis is essentially an assumption that leads into our logic more formally. The role of this step is often misunderstood, but here's how it works in a simple manner:

• Assume that the equation holds true for some arbitrary positive integer \( k \). In our problem, this means accepting that \( 2 + 4 + 8 + \dots + 2^k = 2^{k+1} - 2 \) is correct under the assumption, without immediate proof for now.
• It's this assumption that will form the pivot point for demonstrating the statement for the next integer.

This step doesn’t prove anything by itself, but it lays the groundwork for making logical connections in the subsequent step.

Induction Step

This step is the powerhouse of mathematical induction. The induction step leverages the hypothesis from the previous section to extend the truth to the next integer, \( n = k + 1 \). Here’s how it’s done:

• We add the term for \( n = k + 1 \), which is \( 2^{k+1} \), to both sides of our assumed equation: \( 2 + 4 + 8 + \dots + 2^k + 2^{k+1} \).
• This results in altering the right side to \( 2^{k+1} - 2 + 2^{k+1} = 2 \times 2^{k+1} - 2 \).
• Simplifying this shows that indeed it holds: \( 2^{k+2} - 2 \), aligning the left and the newly formed right side.

Completing the induction step verifies that if true for \( n = k \), our statement is also valid for \( n = k + 1 \), effectively linked each step to the next.

Positive Integers

Mathematical induction applies to a special set of numbers called positive integers, or the set starting from 1 and counting upward (\( 1, 2, 3, \ldots \)). In our induction proof, they serve as the values that \( n \) can take.

Understanding positive integers in this context means:

• The base case started with \( n = 1 \), rooting our proof at the single smallest positive integer.
• The process of induction demonstrates that since it holds for \( 1 \), it will be true for the entire series of 2, 3, 4, and so on, concretely affirming our statement for all positive integers thereafter.

By spanning the complete range of these integers, mathematical induction reassures us that our equation isn't just incidentally true but rather universally so across the numbers we care about.