Problem 25 - Precalculus: Mathematics for Calculus

Question

A sequence is defined recursively by \(a_{n+1}=3 a_{n}\) and \(a_{1}=5 .\) Show that \(a_{n}=5 \cdot 3^{n-1}\) for all natural numbers \(n\)

Step-by-Step Solution

Answer

The formula \(a_n = 5 \cdot 3^{n-1}\) is proven by induction to hold for all natural numbers \(n\).

Step 1: Understand the Recursive Sequence

The sequence is defined recursively, which means each term depends on the one before it. It starts with \(a_1 = 5\), and each subsequent term is found using \(a_{n+1} = 3a_n\).

Step 2: Propose a Formula for the Sequence

We propose that the sequence is given by the explicit formula \(a_n = 5 \cdot 3^{n-1}\). Our task is to show this formula holds for all natural numbers \(n\).

Step 3: Base Case Verification

First, check the base case for \(n = 1\). The recursive definition gives \(a_1 = 5\). By the proposed formula, \(a_1 = 5 \cdot 3^{1-1} = 5\cdot 3^0 = 5\cdot 1 = 5\). Thus, the formula holds for \(n = 1\).

Step 4: Inductive Step Setup

Assume the formula is true for \(n = k\), i.e., \(a_k = 5 \cdot 3^{k-1}\). We need to show it holds for \(n = k+1\).

Step 5: Use Inductive Hypothesis for \(n=k+1\)

Based on the recursive formula, \(a_{k+1} = 3a_k\). By the induction hypothesis, substitute \(a_k = 5 \cdot 3^{k-1}\) into the equation: \(a_{k+1} = 3 \times 5 \cdot 3^{k-1} = 5\cdot 3\cdot 3^{k-1}\).

Step 6: Simplify to Match Proposed Formula

Simplify the expression for \(a_{k+1}\):\[5 \cdot 3 \cdot 3^{k-1} = 5 \cdot 3^1 \cdot 3^{k-1} = 5 \cdot 3^{1+k-1} = 5 \cdot 3^k\]This matches the proposed formula for \(n = k+1\), which is \(a_{k+1} = 5 \cdot 3^k\).

Step 7: Conclusion of Induction Proof

Since the formula is true for \(n = 1\) (base case) and holds from \(n = k\) to \(n = k+1\), by induction, the formula \(a_n = 5 \cdot 3^{n-1}\) is true for all natural numbers \(n\).

Key Concepts

Mathematical Induction

Mathematical induction is an essential method in mathematics, often used to prove statements for all natural numbers. It's like verifying that dominoes in a row will fall

• First, confirm the first domino falls.
• Next, show that if one domino falls, the next will inevitably fall too.

In mathematical terms, we establish a base case to show the formula works for the initial number. Then, using the induction hypothesis, you assume it’s true for a particular case and prove it for the next. The proof involves showing the bridge between one domino and the next is sound. If both parts are confirmed, your formula or statement holds true for all dominoes—or in this case, all natural numbers.

Explicit Formula

When sequences are defined, they can be recursive like our initial sequence, or explicit. An explicit formula allows you to directly compute the value of any term in a sequence without referencing any of its predecessors.

• For our sequence, the explicit formula is given as \( a_n = 5 \cdot 3^{n-1} \).
• This means, for any term number \( n \), we can easily find the term using this formula.

Unlike recursive formulas, explicit ones provide a direct computation path. They look at the broader pattern and generalize it into a single expression, which is much quicker and more efficient for larger sequences. Hence, transforming a recursive sequence into an explicit formula makes calculations straightforward and readily accessible.

Base Case

The base case is the foundational step in mathematical induction. This step is crucial as it proves the formula holds for the smallest value of \( n \), typically \( n=1 \). By verifying:

• With \( a_1 = 5 \cdot 3^{1-1} \), the given base formula results in \( 5 \cdot 1 = 5 \).
• It confirms the formula's validity for the first term, matching the recursive definition of \( a_1 = 5 \).

Once the base case is verified, it offers a starting anchor point, ensuring that the statement being proven is true from the very beginning of the sequence. Thus, providing a clear beginning for the induction process to build upon.

Inductive Step

To complete the process of mathematical induction, the inductive step is essential. Here, you assume the formula is accurate for a particular term \( n=k \), which in our scenario means assuming \( a_k = 5 \cdot 3^{k-1} \).

• Next, show it holds for the next instance, \( n=k+1 \), using the recursive relationship.
• By substituting \( a_k \) into the recursive formula \( a_{k+1} = 3 \cdot a_k \), you show \( a_{k+1} = 5 \cdot 3^k \).

The inductive step works as the bridge from one verified case to all subsequent cases. It ensures any case within the sequence leads to the truth of the next, thereby supporting the truth of the formula to extend throughout the entire natural number sequence.