Question

Prove that \(3^{1}+3^{2}+3^{3}+3^{4}+\cdots+3^{n}=\frac{3^{n+1}-3}{2}\) for every \(n \in \mathbb{N}\).

Short answer

The formula \(3^{1}+3^{2}+3^{3}+3^{4}+\cdots+3^{n}=\frac{3^{n+1}-3}{2}\) holds for all natural numbers \(n\).

Step-by-step solution

Base Case

The base case for \(n\) is 1. Substitute \(n = 1\) into the two sides of the equation to verify this. On the left side, the sum becomes \(3^{1}\) which equals to 3. On the right side, we have \(\frac{3^{1+1}-3}{2} = \frac{3^{2}-3}{2} = \frac{9-3}{2} = 3\). This means the formula holds for \(n = 1\).

Inductive Hypothesis

Assume the formula is true for some positive integer \(n = k\), that is \(3^{1}+3^{2}+3^{3}+3^{4}+\cdots+3^{k}=\frac{3^{k+1}-3}{2}\)

Inductive Step

We need to show that the formula is true for \(n = k+1\), meaning \(3^{1}+3^{2}+3^{3}+3^{4}+\cdots+3^{k} + 3^{k+1}=\frac{3^{k+2}-3}{2}\). Add \(3^{k+1}\) to both sides of our inductive hypothesis. On left side, it becomes the sum up to \(k+1\). On the right side, we get \(\frac{3^{k+1}-3}{2} + 3^{k+1} = \frac{2*(3^{k+1})-3}{2} = \frac{3^{k+2}-3}{2}\). This proves the formula holds for \(n = k+1\) when it is true for \(n = k\). Therefore, the formula is true for all \(n \in \mathbb{N}\).

Key concepts

Proof Theory

Proof theory is a fundamental area within mathematical logic that examines the nature, structure, and prospects of mathematical proof. It aims to provide a formal basis for determining the validity of mathematical statements. In the context of our exercise involving mathematical induction, proof theory instills a framework for us to establish the truth of an infinite number of cases with a finite number of steps.

Proofs by induction, such as the one given in our exercise, follow a two-step process: the base case and the inductive step. Beginning with confirming the statement for an initial case—usually for the smallest value in the domain, which for natural numbers is often one. After the base case is verified, the inductive step assumes the statement holds for an arbitrary case 'k' and then shows that it must also hold for 'k+1'. If both these steps are successfully shown, then proof theory allows us to conclude that the statement is true for all natural numbers.

Natural Numbers

Natural numbers are the set of all positive integers starting from 1, moving upwards without limit. They are the numbers we naturally visualize when we count objects, and in mathematics, they are denoted by \( \mathbb{N} \). In our exercise, the variable \( n \) is considered to be a member of this infinite set.

The property of being unbounded makes natural numbers ideal for concepts involving growth or sequences, and it's a key part of why mathematical induction is so powerful—it allows us to prove statements for a potentially infinite sequence of natural numbers with just a few logical steps. Whenever we see variables that belong to the set of natural numbers, we're often thinking about sequences, series, counting, and occasionally, proving statements via induction as we do in our exercise.

Exponential Series

Exponential series refer to sequences where each term is an exponentiation of a constant basis with increasing powers. In our solved problem, the base is 3, and the exponents range from 1 up to some natural number \( n \), creating the series \(3^{1}+3^{2}+3^{3}+3^{4}+\cdots+3^{n}\).

The series amplifies rapidly due to the exponential nature of each term. Understanding the summation of such a series is valuable in various mathematical and real-world applications, such as computing compound interest or modeling population growth. The formula provided in the solution, \(\frac{3^{n+1}-3}{2}\), elegantly summarizes the total of the exponential series in a way that's much simpler to calculate compared to adding up potentially hundreds or even an infinite number of terms. This type of reduction from a complex series to a single expression is a powerful technique in mathematics, making daunting calculations both understandable and feasible.