Question

The terms of a sequence are given recursively as \(a_{0}=2, a_{1}=6,\) and \(a_{n}=2 a_{n-1}+\) \(3 a_{n-2}\) for \(n \geq 2\). Prove by induction that \(b_{n}=2 \cdot 3^{n}\) is a closed form for the sequence.

Short answer

The sequence \( a_n \) has the closed form \( b_n = 2 \cdot 3^n \).

Step-by-step solution

Base Case Verification

Check if the closed formula holds true for the initial terms, specifically for \( n=0 \) and \( n=1 \).For \( n=0 \), the closed form is \( b_0 = 2 \cdot 3^0 = 2 \). This matches \( a_0 = 2 \).For \( n=1 \), the closed form is \( b_1 = 2 \cdot 3^1 = 6 \). This matches \( a_1 = 6 \).The base cases hold true for both \( n=0 \) and \( n=1 \).

Inductive Hypothesis

Assume that the closed formula \( b_k = 2 \cdot 3^k \) is true for some arbitrary \( k \geq 1 \). In mathematical terms, suppose \( b_k = 2 \cdot 3^k = a_k \) and \( b_{k-1} = 2 \cdot 3^{k-1} = a_{k-1} \).

Inductive Step

Using the inductive hypothesis, demonstrate the closed form for \( n = k+1 \).The recursive formula is given by:\[ a_{k+1} = 2a_k + 3a_{k-1} \]Substitute the inductive hypothesis into the equation:\[ a_{k+1} = 2(2 \cdot 3^k) + 3(2 \cdot 3^{k-1}) \]Simplify the expression:\[ a_{k+1} = 4 \cdot 3^k + 6 \cdot 3^{k-1} = 4 \cdot 3^k + 2 \cdot 3^k \]Combine like terms:\[ a_{k+1} = 6 \cdot 3^k = 2 \cdot 3^{k+1} \] which matches \( b_{k+1} \).

Conclusion

Since the base cases hold, and if the statement is true for some \( k \), it is also true for \( k+1 \), by the principle of mathematical induction, \( b_n = 2 \cdot 3^n \) is the closed form of the sequence.

Key concepts

Recursive Sequences

A recursive sequence is one in which each term is defined based on previous terms. It's like building a tower where each new level relies on the ones below it. For example, consider the sequence in the original exercise:

• The sequence starts with initial terms: \( a_0 = 2 \) and \( a_1 = 6 \).
• For any term \( a_n \) where \( n \geq 2 \), it is calculated using the formula: \( a_n = 2a_{n-1} + 3a_{n-2} \).

Each term depends on the two preceding terms. This kind of definition helps in modeling situations where the present depends on the history, such as population growth or financial interests.
Using recursive sequences allows us to systematically predict future terms using simpler base rules without directly calculating each term anew.

Closed Form

A closed form expression allows us to find terms without needing to reference previous terms. It's like having a shortcut, where you don't need to travel the whole journey again to reach a destination. In the original exercise, proving that \( b_n = 2 \cdot 3^n \) is a closed form for the recursive sequence is crucial.

• This expression gives us a direct way to calculate the \( n \)-th term based on \( n \), without piecing through all preceding terms.
• A closed form is desirable because it simplifies understanding and allows for quick computation of terms.

By converting a recursive sequence into a closed form, solving complex problems becomes a matter of simple arithmetic rather than iterative calculations.

Base Case

The base case is the starting point in mathematical induction where we verify the statement for the initial term or terms. It's the foundation that helps ensure the entire proof is rooted in truth.
In this exercise:

• We checked the base cases at \( n=0 \) and \( n=1 \).
• Both cases matched perfectly with the recursive equation, verifying the closed form holds for these initial values.

By confirming these base cases, we set the stage for proving the pattern holds true for all subsequent terms using mathematical induction.

Inductive Hypothesis

The inductive hypothesis is the assumption that forms the bridge between the known base cases and proving the statement for all terms. It's a temporary acceptance of the truth of the statement for some arbitrary value.
In this step:

• We assume the closed form \( b_k = 2 \cdot 3^k \) is true for some \( k \geq 1 \).
• This assumption helps in proving that if it holds for this \( k \), it should also hold for \( k+1 \).

Constructing this bridge via the inductive hypothesis is essential in mathematical induction. It allows us to extend the validity of the base case to every term in the sequence, step by step, without having to prove each one individually.