Problem 23 - 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 \(a_n = 5 \cdot 3^{n-1}\) holds by mathematical induction.

Step 1: Understand the given recurrence relation

We are given that the sequence follows the recurrence relation \(a_{n+1} = 3a_{n}\) and that the first term \(a_1 = 5\). This means starting from \(a_1\), each subsequent term is obtained by multiplying the previous term by 3.

Step 2: Verify initial terms match the proposed formula

According to the formula \(a_n = 5 \cdot 3^{n-1}\), let’s verify this for early values of \(n\): For \(n=1\), \(a_1 = 5 \cdot 3^{1-1} = 5 \cdot 1 = 5\). This matches the initial condition. For \(n=2\), \(a_2 = 5 \cdot 3^{2-1} = 5 \cdot 3 = 15\). This should match \(3a_1 = 3 \cdot 5 = 15\).For \(n=3\), \(a_3 = 5 \cdot 3^{3-1} = 5 \cdot 9 = 45\). This should match \(3a_2 = 3 \cdot 15 = 45\).These match the given recurrence relation.

Step 3: Use mathematical induction for general proof

We'll use mathematical induction to prove that \(a_n = 5 \cdot 3^{n-1}\) holds for all natural numbers \(n\):**Base Case**: For \(n = 1\), \(a_1 = 5\ \cdot 3^{1-1} = 5\cdot 1 = 5\) which is true by initial condition.**Inductive Step**: Assume \(a_k = 5 \cdot 3^{k-1}\) is true for some \(k\). We need to show it is true for \(k+1\). Starting from the inductive hypothesis, \(a_{k+1} = 3a_k = 3(5 \cdot 3^{k-1}) = 5 \cdot 3^k\).Thus, the formula holds for \(k+1\). Hence, by induction, \(a_n = 5 \cdot 3^{n-1}\) is valid for all \(n\).

Key Concepts

Mathematical Induction

Mathematical induction is a powerful technique used to prove statements or formulas that are asserted to be true for all natural numbers. The process involves two key steps:

• Base Case: Start by proving the statement for the initial value, usually when n equals 1. This sets the foundation for the proof.
• Inductive Step: Assume that the statement holds for an arbitrary natural number k. Then, prove that if the statement is true for k, it must also be true for k+1. This step ensures that the truth of the statement "carries forward" to all subsequent numbers.

In our exercise, we start with the base case verification for n=1, showing that the initial term matches the proposed formula: 5. The inductive hypothesis assumes that our statement is true for some natural number k. By proving it holds for k+1 as well, we establish that the formula is universally correct for all values of n. This method is crucial for validating statements, especially in sequences and series.

Sequence and Series

Sequences and series are foundational concepts in mathematics that describe ordered lists of numbers. A sequence is simply a list of numbers in a specific order, while a series is the sum of a sequence of numbers.

• Sequence: In a sequence, each number is called a term, and sequences can be finite or infinite. They are usually defined by a specific formula or a recurrence relation, like in our given problem where each term depends on its predecessor.
• Series: When you sum the terms of a sequence, you create a series. A series can also be finite or infinite, such as an arithmetic or geometric series.

For the given exercise, the sequence is defined recursively by the relation \(a_{n+1} = 3a_{n}\), with an initial term \(a_1 = 5\). This setup forms a geometric sequence where each term is a product of the previous one and a constant ratio. Understanding sequences helps to analyze various mathematical phenomena, modeling natural patterns and solving complex problems.

Geometric Progression

A geometric progression, or geometric sequence, is a sequence where each term after the first is determined by multiplying the previous one by a fixed, non-zero number known as the common ratio. The general form of a geometric progression can be expressed as:

• Starting term: \(a_1\)
• Common ratio: \(r\)
• General term formula: \(a_n = a_1 \cdot r^{n-1}\)

In the context of our problem, the sequence \(a_n\) is a geometric progression where the initial term \(a_1 = 5\) and the common ratio \(r = 3\). The derived formula \(a_n = 5 \cdot 3^{n-1}\) confirms this pattern, nicely illustrating how each term builds upon the previous one following the constant geometric rule. Understanding geometric progressions is essential for solving real-world problems in finance, physics, and other fields where exponential growth or decay is observed.