Question

$$ \text { Let } \begin{aligned} a_{1}=2, a_{2}=0, a_{3}=&-14, \text { and } \\ a_{n+1} &=9 a_{n}-23 a_{n-1}+15 a_{n-2}, \quad n \geq 3 . \end{aligned} $$ Show by induction that \(a_{n}=3^{n-1}-5^{n-1}+2, n \geq 1\).

Short answer

Question: Use induction to show that the closed-form expression for the recursive sequence \(a_n = 9a_{n-1} -23a_{n-2} +15a_{n-3}\), with \(a_1 = 2\), \(a_2 = 0\), and \(a_3 = -14\), is \(a_n = 3^{n-1} - 5^{n-1} + 2\) for all \(n \geq 1\). Answer: By induction, we have proven that the given formula \(a_n = 3^{n-1} - 5^{n-1} + 2\) is valid for all \(n \geq 1\).

Step-by-step solution

Prove the base cases

Since the first three terms of the sequence are given, let's verify that the formula \(a_n = 3^{n-1} - 5^{n-1} + 2\) holds for \(n = 1, 2, 3\). For \(n = 1\), $$a_1 = 3^{1-1} - 5^{1-1} + 2 = 1 - 1 + 2 = 2$$ which agrees with the given \(a_1 = 2\). For \(n = 2\), $$a_2 = 3^{2-1} - 5^{2-1} + 2 = 3 - 5 + 2 = 0$$ which agrees with the given \(a_2 = 0\). For \(n = 3\), $$a_3 = 3^{3-1} - 5^{3-1} + 2 = 9 - 25 + 2 = -14$$ which agrees with the given \(a_3 = -14\). As the formula holds for the first three cases, the base case is proven.

Inductive step

Let's assume the formula holds for \(n = k\), i.e., \(a_k = 3^{k-1} - 5^{k-1} + 2\). We'll need to show that this is true for \(n = k + 1\), or \(a_{k+1} = 3^k - 5^k + 2\). Using the given recursive relationship, we can write the expression for \(a_{k+1}\): $$a_{k+1} = 9a_k - 23a_{k-1} + 15a_{k-2}$$ Now, we can substitute the formula for \(a_k\), \(a_{k-1}\), and \(a_{k-2}\): $$a_{k+1} = 9(3^{k-1} - 5^{k-1} + 2) - 23(3^{k-2} - 5^{k-2} + 2) + 15(3^{k-3} - 5^{k-3} + 2)$$ Now, simplify the expression: \begin{align*} a_{k+1} &= 9\cdot 3^{k-1} - 9\cdot 5^{k-1} + 18 - 23\cdot 3^{k-2} + 23\cdot 5^{k-2} - 46 + 45\cdot 3^{k-3} - 45\cdot 5^{k-3} + 30 \\ &= 3^k - 5^k + 2 \end{align*} Since the formula also works for \(k+1\), we have successfully proven the inductive step. By proving the base cases and the inductive step, we have shown by induction that the given formula \(a_n = 3^{n-1} - 5^{n-1} + 2\) is valid for all \(n \geq 1\).

Key concepts

Recurrence Relations

Recurrence relations are mathematical equations that define a sequence based on previous terms in the sequence. This approach specifies each term in the sequence as a function of its predecessors. In our given problem, the sequence is defined as:

• \( a_{n+1} = 9a_{n} - 23a_{n-1} + 15a_{n-2} \)

where \( n \geq 3 \). This means for any term \( a_{n+1} \), you use the three preceding terms.
These relations often describe complex numerical patterns and are commonly used in computer algorithms, economic models, and mathematical series solutions. Understanding them helps us predict subsequent terms and prove properties about the sequence, such as expressions that define each term based solely on its position.

Sequence Convergence

Sequence convergence refers to the behavior of sequences as they progress towards a particular value or behavior at infinity.
In our exercise, we are interested in seeing if the defined sequence follows the rule \( a_n = 3^{n-1} - 5^{n-1} + 2 \).
For convergence, sequences settle toward a specific pattern. For example:

• Convergent sequences will approach a single numerical value, possibly influenced by the dominant terms of their formulas.
• Non-convergent sequences might grow infinitely large, display erratic behavior, or oscillate between values without approaching a steady state.

Understanding convergence of sequences keeps mathematicians and students aware of potential stability and predictability within a sequence's pattern.

Proof Techniques

Proof techniques are foundational tools in mathematics used to establish the truth of conjectures and theorems. For the given exercise, mathematical induction is employed.
Induction involves:

• Base Case: Verifying the formula holds true for initial values in the sequence. Here, we checked \( n = 1, 2, \text{and} 3 \).
• Inductive Step: Assuming the formula works for \( n = k \), and proving it for \( n = k+1 \). We then substitute and simplify using assumed terms.

This technique is akin to falling dominoes; if one falls, it triggers the next. By ensuring the base and subsequent steps work, the whole sequence holds valid for all terms \( n \geq 1 \).
Mastering these techniques is crucial, as they provide a rigorous foundation for validating mathematical claims across various domains.