Question

Prove the following statements with either induction, strong induction or proof by smallest counterexample. If \(n \in \mathbb{N}\), then \(2^{1}+2^{2}+2^{3}+\cdots+2^{n}=2^{n+1}-2\).

Short answer

For \(n=2\), the equation holds as the base case. Assuming that the equation is true for \(n=k\), it can be shown that the equation also holds for \(n=k+1\), completing the induction proof. Therefore, the given equation is true for all natural numbers \(n \geq 2\).

Step-by-step solution

Base Case

Check if the equation holds for \(n=1\). On the LHS we have \(2^{1} = 2\), and on the RHS we have \(2^{1+1} - 2 = 2 - 2 = 0\). Hence, it does not hold for \(n=1\). Now let's try \(n=2\). On the LHS we have \(2^{1} + 2^{2} = 2 + 4 = 6\), and on the RHS we get \(2^{2+1} - 2 = 8 - 2 = 6\). Hence, the base case is \(n=2\).

Induction Hypothesis

Assume that the given equation is true for \(n=k\), i.e., \(2^{1}+2^{2}+\cdots+2^{k}=2^{k+1}-2\). This is called the induction hypothesis.

Induction Step

Prove the equation for \(n=k+1\) under the assumption that it holds for \(n=k\). Write out the LHS for \(n=k+1\): \(2^{1}+2^{2}+\cdots+2^{k}+2^{k+1}\). Substitute the induction hypothesis into the LHS, which gives: \((2^{k+1}-2) + 2^{k+1} = 2 \cdot 2^{k+1}- 2 = 2^{k+2}-2\). The RHS for \(n=k+1\) is \(2^{k+2}-2\), so this shows that the equation holds for \(n=k+1\) if it holds for \(n=k\).

Key concepts

Mathematical Induction

Mathematical induction is a powerful proof technique often likened to a domino effect; to prove that a property holds for all members of a well-ordered set, usually the natural numbers.

Imagine lining up infinite dominoes. If you can 1) Prove the first one falls down (the base case) and 2) Show that when any domino falls, it guarantees the fall of the next one (the induction step), then all the dominos will fall.

• The base case confirms the statement is true for the initial value.
• The induction step proves that if the statement holds for an arbitrary instance, it also holds for the next instance.

By proving these two conditions, you can conclude the statement is valid for all natural numbers.

Strong Induction

Strong induction, closely related to simple mathematical induction, often comes into play when the induction step requires not just the immediately preceding instance but possibly several previous instances or even all instances up to the current point.

While the principle is the same – proving a base case and an induction step – the strength comes from the broader assumption. Rather than assuming the statement holds for just one preceding case, it hypothesizes the truth for all cases leading up to and including a particular instance, allowing you to prove the statement for the next instance.

• This technique is particularly useful when each step relies on multiple predecessors rather than just one.

Proof by Smallest Counterexample

Proof by smallest counterexample is a contradiction-based strategy. The aim is to assume that the statement in question is false, and there exists a smallest counterexample which disproves the statement.

Then, we derive a contradiction by showing the existence of an even smaller counterexample, which is impossible because we stated we had the 'smallest' one. This contradiction implies that no such smallest counterexample can exist, thus the original statement must be true.

• This approach can often reveal the underlying patterns or properties that rigorous induction methods can then use to form a robust proof.

Base Case

In the context of the provided exercise, establishing the base case is the cornerstone of mathematical induction. Here, we must determine the initial instance where the given statement holds true, which will kickstart the induction process.

While often this means starting the proof with n=1, we have to be careful. The key is to find the correct first instance, and for some sequences or formulas, the base case might not be the first natural number.

• In this exercise, the correct base case is found to be when n=2, which guides us to ensure that our logic is tailored to the specifics of the problem at hand.

Induction Hypothesis

The induction hypothesis is a critical assumption during the induction step. In the exercise, by assuming that our proposition holds at an arbitrary stage k, we leverage this belief to prove the property for the subsequent stage k+1.

It is essential to understand that the induction hypothesis isn't about proving the statement true for k; it's assumed true as a means to prove it for k+1.

• Properly applying the induction hypothesis involves a careful and sometimes creative manipulation of the equation or inequality to show that the truth of the statement naturally extends to the next case in the sequence.