Problem 19 - Algebra and Trigonometry

Question

Use mathematical induction to prove that each statement is true for every positive integer n. \(2+4+8+\cdots+2^{n}=2^{n+1}-2\)

Step-by-Step Solution

Answer

By applying the principle of mathematical induction, the given statement is proven to be true for all \(n\).

Step 1: Base Case

Prove the statement is true for \(n=1\). Indeed, the left side is \(2\) and the right side is \(2^{1+1}-2=2\). Hence, the statement holds for \(n=1\).

Step 2: Induction Step

Assume that the statement is true for some \(k \geq 1\), that is, \(2+4+8+\cdots+2^{k}=2^{k+1}-2\).

Step 3: Prove the statement for \(k+1\)

We need to show that the statement holds for \(k+1\). Consider the left side, which is the sum of the first \(k+1\) terms. It can be written as \( (2+4+8+\cdots+2^{k}) + 2^{k+1}\). By our induction hypothesis, this is equal to \(2^{k+1}-2 + 2^{k+1} = 2\cdot2^{k+1}-2 = 2^{k+2}-2\), which equals to the right hand side if \(n\) is replaced by \(k+1\). Therefore, the formula holds for \(k+1\) when it holds for \(k\), proving our induction step.

Step 4: Conclusion

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

Key Concepts

Understanding Proof by Induction

Proof by induction is a mathematical technique used to prove that a proposition or statement is true for all natural numbers. It's akin to a domino effect; if you can knock over the first one and prove that knocking one down causes the next to follow, they will all fall in sequence.

The process involves two main steps. The first is the 'base case', where the statement is proven true for an initial value, typically for the smallest value for which the statement is claimed to be true. In the exercise we're exploring, the base case is when the value of 'n' is 1. Here, both sides of the equation amount to the same value, which satisfactorily checks off our first step.

The second step is the 'induction step'. Assuming the statement holds for some arbitrary positive integer 'k', we then show it also holds for 'k+1'. If successful, this means the statement is true for 'k+2', 'k+3', and so on. This is comparable to showing that the domino effect continues without end. In our exercise, this continuation is confirmed by manipulating the sum of the series up to the 'k+1'th term and showing it aligns with the expected result on the right side of the equation.

The Sum of a 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 fixed, non-zero number called the common ratio. For example, in the sequence 2, 4, 8, and so on, each term is double the preceding one, making this a geometric series with a common ratio of 2.

In algebra, the sum of the first 'n' terms of a geometric series can be expressed using a formula. For our series with a common ratio of 2, the sum up to the 'n'th term is represented as the sum of a sequence of exponents of 2.

Deriving the Sum

By employing the formula for the sum of a geometric series, we can find a neat expression for the sum of our series. The exercise demonstrates this by showing that the sum of the series up to the 'n'th term is equal to \(2^{n+1}-2\). This result is key to understanding how the induction proof is structured and executed.

Working with Exponents

Exponents represent how many times a number, known as the base, is multiplied by itself. For example, \(2^n\) indicates that 2 is multiplied by itself 'n' times. It is a fundamental concept in mathematics, especially when dealing with geometric series as it frequently describes the terms in the series.

Exponents in the Geometric Series

In our exercise, each term of the series can be represented as \(2^k\), where 'k' ranges from 1 to 'n'. It's pivotal in the induction step to show how adding the next term \(2^{k+1}\) to the assumed sum \(2^{k+1}-2\) leads to \(2^{k+2}-2\), elegantly preserving the pattern. Grasping exponent rules, such as \(2^{k+1} \cdot 2 = 2^{k+2}\), is essential in concluding the proof. It illuminates why each step in the induction reflects consistent exponential growth, thereby further cementing the underlying concept of the geometric progression.