Problem 25 - Algebra and Trigonometry

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 formula \( a_n = 5 \cdot 3^{n-1} \) is valid for all natural numbers \( n \).

Step 1: Understand the Problem

We need to find the general formula for the sequence defined by the recurrence relation \( a_{n+1} = 3a_n \) with the initial condition \( a_1 = 5 \), and prove that the formula \( a_n = 5 \cdot 3^{n-1} \) holds for all natural numbers \( n \).

Step 2: Base Case Verification

Start by checking the base case when \( n = 1 \). Since \( a_1 = 5 \), substitute \( n = 1 \) into the formula: \( a_n = 5 \cdot 3^{1-1} = 5 \). Thus, the formula holds for \( n = 1 \).

Step 3: Inductive Hypothesis

Assume that the formula holds for some natural number \( k \); that is, assume \( a_k = 5 \cdot 3^{k-1} \). This assumption is our inductive hypothesis.

Step 4: Inductive Step

We need to show that if the formula holds for \( n = k \), it also holds for \( n = k+1 \). According to the recurrence relation, \( a_{k+1} = 3a_k \). Substitute the inductive hypothesis: \( a_{k+1} = 3(5 \cdot 3^{k-1}) = 5 \cdot 3^k \). Thus, the formula holds for \( n = k+1 \).

Step 5: Conclusion

By the principle of mathematical induction, since the formula holds for \( n = 1 \) and \( n = k+1 \) whenever it holds for \( n = k \), the formula \( a_n = 5 \cdot 3^{n-1} \) is true for all natural numbers \( n \).

Key Concepts

Recurrence Relations

When studying sequences in mathematics, recurrence relations are essential tools. A recurrence relation, in simple terms, defines each term in the sequence using the previous terms. This concept helps us understand how sequences evolve over time. In our exercise, the recurrence relation is given as:

• \( a_{n+1} = 3a_n \) for all natural numbers \( n \)

Here, each term, \( a_{n+1} \), is three times the previous term, \( a_n \).
This structure tells us how the sequence progresses and gives us a way to predict subsequent terms, provided we know the initial term, which in this case is given as \( a_1 = 5 \).
Recurrence relations are powerful because they translate a sequence into a more understandable, step-by-step format.

General Formula

A general formula provides an explicit expression to find any term in a sequence without calculating all previous terms. For our problem, the challenge is to derive a formula for \( a_n \) using the recurrence relation.

• The supposed solution is \( a_n = 5 \cdot 3^{n-1} \).

In formal terms, the general formula simplifies finding a sequence's terms by plugging \( n \) directly into the equation.
Instead of iteratively applying the recurrence relation, the general formula makes the sequence easily manageable and calculable.
By using the general formula, students can understand both the pattern within the sequence and efficiently compute any term they need. Understanding this formula is crucial, as it reflects how the sequence grows based on both the initial value and the multiplying factor (in this case, 3).

Base Case

In mathematical induction, the base case is the initial step where we verify the statement for the first element of the set. In simpler terms, we check if our general formula holds true for the first term of the sequence.

• For this exercise, that first element is \( a_1 \), which is given to be 5.

We substitute \( n = 1 \) into the proposed general formula:

• \( a_1 = 5 \cdot 3^{1-1} = 5 \cdot 3^0 = 5 \).

Since this result matches our given \( a_1 = 5 \), the base case verification is successful.
Validating the base case is fundamental. It establishes the starting point from which we can confidently test the hypothesis for all subsequent numbers in our sequence. Without this base, the inductive process would lack foundation.

Inductive Step

The inductive step is a crucial part of mathematical induction. It helps us prove that if our formula is true for one specific case, \( n = k \), it will also be true for the next case, \( n = k+1 \).
In this exercise, the inductive step involves two main parts:

• Assuming \( a_k = 5 \cdot 3^{k-1} \) is true (Inductive hypothesis).
• Using this assumption, we show \( a_{k+1} = 5 \cdot 3^k \).

Through the recurrence relation, we know \( a_{k+1} = 3a_k \).
Substituting the inductive hypothesis gives:

• \( a_{k+1} = 3(5 \cdot 3^{k-1}) = 5 \cdot 3^k \)

This confirms that if the formula holds for \( a_k \), it should also hold for \( a_{k+1} \).
The inductive step ensures that our general formula applies universally across the sequence, asserting the correctness for each natural number. It essentially "walks" the truth of our formula from one term to the next, establishing a continuous chain of validation.