Problem 25 - Precalculus Mathematics for Calculus

Question

Let \(a_{n+1}=3 a_{n}\) and \(a_{1}=5 .\) Show that \(a_{n}=5 \cdot 3^{n-1}\) for all natural numbers \(n\)

Step-by-Step Solution

Answer

The sequence formula is \(a_n = 5 \cdot 3^{n-1}\) for all \(n\).

Step 1: Understand the Initial Conditions

To solve the problem, first understand that you are given a recursive relationship: \(a_{n+1} = 3a_n\). This means to find any term in the sequence, you multiply the previous term by 3. Also, the first term of the sequence is \(a_1 = 5\).

Step 2: Analyze the Given Formula

The problem states that we should show \(a_n = 5 \cdot 3^{n-1}\). This formula gives an explicit way to calculate the sequence without using recursion. Our job is to verify that this formula holds true for all natural numbers \(n\).

Step 3: Verify the Formula for the Base Case

Verify the formula for the base case where \(n = 1\). Substituting 1 into the explicit formula gives:\[ a_1 = 5 \cdot 3^{1-1} = 5 \cdot 3^0 = 5 \cdot 1 = 5. \] This matches the initial condition \(a_1 = 5\), confirming that the formula works for the base case.

Step 4: Inductive Step Setup

Assume that the formula \(a_k = 5 \cdot 3^{k-1}\) is true for some arbitrary natural number \(k\). This is called the inductive hypothesis.

Step 5: Inductive Step: Show the Next Case

We need to show that if \(a_k = 5 \cdot 3^{k-1}\) is true, then \(a_{k+1} = 5 \cdot 3^k\) is also true. Start by using the recursive relationship: \(a_{k+1} = 3a_k\).

Step 6: Perform Inductive Calculation

Substitute the inductive hypothesis into the recursive formula:\[ a_{k+1} = 3a_k = 3(5 \cdot 3^{k-1}) = 5 \cdot 3 \cdot 3^{k-1} = 5 \cdot 3^k. \] This shows that if the formula holds for \(k\), it also holds for \(k+1\).

Step 7: Complete Inductive Proof

Since both the base case and the inductive step have been proved, by mathematical induction, the formula \(a_n = 5 \cdot 3^{n-1}\) holds true for all natural numbers \(n\).

Key Concepts

Recursive Sequences

When dealing with recursive sequences, we focus on using a previous term to determine the next one. This sequence is defined by a specific relationship between consecutive terms. For example, in this exercise, the sequence is defined by the equation \(a_{n+1} = 3a_n\), which means each term depends on the one before it. Starting from an initial term, \(a_1 = 5\), you can find subsequent terms by applying this rule repeatedly.

• The base case is crucial: it sets the stage for the pattern to emerge. Here, our starting point is \(a_1 = 5\).
• Every application of the recursive rule builds on the prior outcome, linking terms through multiplication by 3.

Thus, recursive sequences are like a chain, where each link depends on the one preceding it.

Explicit Formula

An explicit formula allows us to compute any term in a sequence without referring to its predecessors. It provides a direct expression to calculate sequence terms. In our problem, the explicit formula is \(a_n = 5 \cdot 3^{n-1}\).

• This formula is valuable because it eliminates the need for recursion, thus simplifying complex calculations.
• It reveals the intrinsic pattern governing the sequence, offering insights into its behavior without calculation of each step.

The formula \(a_n = 5 \cdot 3^{n-1}\) lets you find any term by simply plugging in the position number, \(n\). This converts the seemingly tedious recursive sequence into a straightforward computation.

Inductive Hypothesis

The process of proving statements for every natural number using induction involves the inductive hypothesis.

• The inductive hypothesis assumes the statement or formula is true for some arbitrary term (say \(k\)). For this problem, we assume \(a_k = 5 \cdot 3^{k-1}\) is true.
• The challenge is to show that this truth carries forward, meaning if it's true for \(k\), it should also be true for \(k+1\).

This approach leverages logical progression:

• Prove for a base case (simplest scenario, often \(n=1\)).
• Assume it's true for an arbitrary \(k\) (inductive hypothesis).
• Show this assumption implies truth for the next term \(k+1\).

By effectively proving both cases, we conclude the formula holds for all terms, ensuring the sequence described functions as specified.