Question

Let \(a_{1}=a_{2}=5\) and $$ a_{n+1}=a_{n}+6 a_{n-1}, \quad n \geq 2 $$ Show by induction that \(a_{n}=3^{n}-(-2)^{n}\) if \(n \geq 1\).

Short answer

Question: Prove by induction that the formula aₙ = 3^{n} - (-2)^{n} holds for all n greater than or equal to 1. Answer: By induction, we have proven that the formula aₙ = 3^{n} - (-2)^{n} holds for all n greater than or equal to 1.

Step-by-step solution

Base Case

First, we need to verify the formula for a₁ and a₂. Let's plug n = 1 and n = 2 into our formula. For n = 1, a₁ = 3¹ - (-2)¹ = 3 - (-2) = 5 For n = 2, a₂ = 3² - (-2)² = 9 - 4 = 5 The formula holds for a₁ and a₂.

Inductive Step

Now, we need to form the Inductive Hypothesis. Let's assume that the formula holds for some arbitrary positive integer n, that is: aₙ = 3^{n} - (-2)^{n}

Induction

We need to prove that the formula holds for aₙ₊₁ under the assumption that the Inductive Hypothesis is true. To do this, we can start by rewriting the expression: aₙ₊₁ = aₙ + 6aₙ₋₁ Now, we can use the Inductive Hypothesis to replace aₙ: aₙ₊₁ = (3^{n} - (-2)^{n}) + 6aₙ₋₁ We also need to replace aₙ₋₁ using the Inductive Hypothesis: aₙ₊₁ = (3^{n} - (-2)^{n}) + 6(3^{n - 1} - (-2)^{n - 1}) Now, let's try to simplify the expression: aₙ₊₁ = 3^{n} - (-2)^{n} + 6 * 3^{n - 1} - 6 * (-2)^{n - 1} To prove the formula, we need to show that this expression is equal to 3^{n + 1} - (-2)^{n + 1}. We can rewrite this expression as: aₙ₊₁ = 3 * 3^{n} - (-2) * (-2)^{n} Now we need to make both expressions look the same: aₙ₊₁ = 3 * 3^{n} - (-2)^{n + 1} + (-2)^{n + 1} - (-2) * (-2)^{n} aₙ₊₁ = 3 * 3^{n} + 2 * (-2)^{n} - (-2)^{n + 1} aₙ₊₁ = 3^{n + 1} - (-2)^{n + 1} Since the two expressions are the same, we have proven that the formula holds for aₙ₊₁ under the assumption that the Inductive Hypothesis is true. Therefore, by induction, the formula a_{n}=3^{n}-(-2)^{n} holds for all n greater than or equal to 1.

Key concepts

Sequence Definition

In mathematics, a sequence is an ordered list of numbers following a particular pattern. Each number in the sequence is called a term. The relationship between consecutive terms is determined by a specific rule, known as the sequence definition. For example, in an arithmetic sequence, each term is derived by adding a constant value to the previous term.

When it comes to the given exercise, the sequence is defined recursively with initial conditions and a recursion formula. The initial conditions give us the starting points: (a_{1}=a_{2}=5). The recursion formula (a_{n+1}=a_{n}+6a_{n-1}) tells us how to find any subsequent term from its preceding terms. Understanding this foundational concept is crucial for successfully applying mathematical induction to prove properties of the sequence.

Inductive Step

The inductive step is a pivotal component of mathematical induction, a process of proving that a statement holds true for all natural numbers. In the inductive step, we assume that the statement is true for some arbitrary case n, and then demonstrate that it is also true for n+1. This assumption that the statement is true for some n is known as the inductive hypothesis.

The power of the inductive step lies in its ability to leap from one case to the next. In the context of our sequence example, this involves manipulating the recursive formula based on the inductive hypothesis. By showing that if the relationship holds for (a_{n} = 3^{n} - (-2)^{n}), then it must also be true for (a_{n+1}), we can effectively convince ourselves of the statement's validity for all natural numbers in the sequence, thus completing our proof.

Inductive Hypothesis

Within the realm of mathematical induction, the inductive hypothesis is a supposition we make stating that the property we aim to prove holds true up to some arbitrary natural number n. It serves as the cornerstone for the inductive step because we use it as a 'stepping stone' to prove that the property then holds for n+1.

In the exercise, the inductive hypothesis is the assumption (a_{n} = 3^{n} - (-2)^{n}) for an unspecified, but specific, positive integer n. By presuming this hypothesis, we empower ourselves to tackle the next term of the sequence, (a_{n+1}), thereby extending the veracity of the formula one number further. The inductive hypothesis is an integral logic tool, bridging our known base case to the broader truth of the mathematical statement for all terms of the sequence.