Question

Given that \(b_{n-1}=2 \cdot 3^{n-1}\) and \(b_{n-2}=2 \cdot 3^{n-2},\) prove that if \(b_{n}=2 b_{n-1}+3 b_{n-2}\) then \(b_{n}=2 \cdot 3^{n}\) provided \(n \geq 2\).

Short answer

The proof shows that \(b_n = 2 \cdot 3^n\) holds for \(n \geq 2\).

Step-by-step solution

Plug in expressions for previous terms

Start by substituting the given expressions for \(b_{n-1}\) and \(b_{n-2}\) into the relation \(b_n = 2b_{n-1} + 3b_{n-2}\). We have:\[ b_{n-1} = 2 \cdot 3^{n-1} \]\[ b_{n-2} = 2 \cdot 3^{n-2} \]. Substitute these into the equation:\[ b_n = 2(2 \cdot 3^{n-1}) + 3(2 \cdot 3^{n-2}) \].

Simplify the equation

Carry out the multiplication and simplify the equation:1. \(2(2 \cdot 3^{n-1}) = 4 \cdot 3^{n-1}\)2. \(3(2 \cdot 3^{n-2}) = 6 \cdot 3^{n-2}\)Now combine these terms:\[ b_n = 4 \cdot 3^{n-1} + 6 \cdot 3^{n-2} \].

Transform into a single power of 3

Express both terms as a single power of 3. Factor out the common term:\[ b_n = (4 \cdot 3^{n-1} + 6 \cdot 3^{n-2}) = 2 \cdot 3^{n-2} (2 \cdot 3 + 3) \].

Simplify the expression further

Simplify the factor inside the parentheses:1. Compute \(2 \cdot 3 = 6\)2. Then compute \(6 + 3 = 9\)Now we have:\[ b_n = 2 \cdot 3^{n-2} \cdot 9 \].

Express as final power of 3

Express 9 as \(3^2\):\[ b_n = 2 \cdot 3^{n-2} \cdot 3^2 \]This simplifies to:\[ b_n = 2 \cdot 3^{n} \].

Conclude the proof

Since we have shown that \( b_n = 2 \cdot 3^n \), we conclude that this holds true for \( n \geq 2 \), completing the proof.

Key concepts

Mathematical Induction

Mathematical induction is a powerful proof technique used to prove that a statement or formula holds true for all natural numbers greater than or equal to a certain value. It works in two steps: the base step and the inductive step.

- **Base Step:** You first prove the statement for the initial value, usually starting at the smallest integer the statement claims to hold. In our problem, we start with proving the statement for the base case, generally at \( n = 2 \) because values at \( n = 0 \) or \( n = 1 \) might not be defined or might be trivial cases.- **Inductive Step:** You assume the statement is true for a particular integer \( k \). This assumption is called the inductive hypothesis. Then you prove the statement is also true for \( k + 1 \). This step bridges the gap, transitioning from one true case to the subsequent one.

If these steps are valid, you show by induction that the statement holds for all numbers greater than the base case. In the provided exercise, mathematical induction would verify that if the formula is true for \( n \), it must also be true for \( n + 1 \).

Sequences

Sequences are ordered lists of numbers following a particular rule or pattern. Understanding sequences is crucial in solving recurrence relations, which describe how each term relates to the previous ones.

In the given exercise, the sequence \( \{b_n\} \) is defined recursively, and the challenge is to express term \( b_n \) in terms of its predecessor terms. Here,- **Recursive Definition:** The expression \( b_n = 2b_{n-1} + 3b_{n-2} \) states each term is related to the two preceding terms. This is helpful when you need to calculate sequential terms without having a closed formula.- **Closed Form:** The solution shows that \( b_n = 2 \cdot 3^n \) represents the closed form of the sequence, offering a direct way to find any term in the sequence without computing all preceding terms.

The closed form gives us a peek into the sequence’s behavior and growth pattern, showing exponential growth driven by the power of 3.

Proof Techniques

Proof techniques are methods used to demonstrate the truth of mathematical statements. In this exercise, we've applied simplification, substitution, and transformation techniques to reveal the underlying structure of the recurrence relation.

- **Substitution:** We substitute the expressions for \( b_{n-1} \) and \( b_{n-2} \) into the recurred equation to rewrite \( b_n \) using known quantities. This brings initial insight into the sequence behavior.- **Simplification:** The next step involves simplifying these expressions, breaking down terms using arithmetic operations. This step helps visualize the sequence in terms of a common base or factor.- **Factoring and Transformation:** The simplification process involves factoring and transforming the expression to reveal a single power in terms of \( 3^n \). This technique cleans up the messiness of prior terms to present a clear expression.

These techniques effectively manage complexity in recurrence relations, providing a way to understand or prove statements within mathematical sequences. By structuring our proofs logically, we build confidence that our conclusions are sound and correct.