Solving the Recurrence Relation ann1(an-1)/3n1: A Comprehensive Guide

Introduction

In the realm of discrete mathematics, recurrence relations are a fundamental concept used to define sequences where each term is defined as a function of the preceding terms. This article will explore the specific recurrence relation given by:

[a_n frac{n1 cdot a_{n-1}}{3^{n1}}]

Our goal is to understand and solve such a recurrence relation, which involves some intricate algebraic steps and potentially requires advanced techniques like mathematical induction. Understanding this will provide insights into how such sequences behave over time, and how one can extract patterns or closed-form solutions.

Understanding the Recurrence Relation

The given recurrence relation can be written as:

[a_n frac{n1 cdot a_{n-1}}{3^{n1}}]

Let's break down the components:

an: The nth term in the sequence. an-1: The (n-1)th term in the sequence, which serves as the base for the current term. n1: A constant factor, which indicates the power to which the base term is multiplied. 3n1: The base term with a constant factor exponentiated.

Factor Out Step

The first step in solving this recurrence relation is to factor out the constant n1 from the sequence definition. Doing so allows us to simplify the recurrence relation and potentially identify a pattern:

[a_n n1 cdot frac{a_{n-1}}{3^{n1}}]

By isolating the constant term, we have:

[a_n n1 left(frac{a_{n-1}}{3^{n1}}right)]

This form reveals that each term in the sequence depends on the previous term multiplied by a factor involving the constant n1 and the exponential term 3n1.

Mathematical Induction Approach

One effective method to solve this recurrence relation is using mathematical induction. This involves proving a statement for the base case and then assuming the statement is true for some arbitrary term and proving it for the next term.

Base Case

Assume we have a starting point, typically denoted as a0, which is a given value or can be determined by the problem context. For simplicity, let's assume:

[a_0 k]

Inductive Step

Assume that the relation holds for an-1. We need to show that it holds for an.

From the recurrence relation, we have:

[a_n n1 left(frac{a_{n-1}}{3^{n1}}right)]

If we assume:

[a_{n-1} k cdot left(frac{1}{3^{n1}}right)^{n-1}]

Then:

[a_n n1 left(frac{k cdot left(frac{1}{3^{n1}}right)^{n-1}}{3^{n1}}right)]

Which simplifies to:

[a_n n1 cdot k cdot left(frac{1}{3^{n1 cdot n}}right)]

We can see that the assumption holds for the next term, confirming the inductive hypothesis.

General Form

Using the mathematical induction approach, we can generalize the form of the sequence. Given the base case a0 k, the general form of the sequence is:

[a_n k cdot n1 cdot left(frac{1}{3^{n1 cdot n}}right)]

This form provides a clear path to understanding the behavior of the sequence as n increases, illustrating how the factor n1 interacts with the exponential decay term (frac{1}{3^{n1 cdot n}}).

Conclusion

Understanding and solving recurrence relations like an n1(an-1)/3n1 is a valuable skill in discrete mathematics and computer science. By breaking down the problem into manageable steps, using techniques like factorization and mathematical induction, we can derive a clear and concise solution.

This article has provided a step-by-step guide to solving such recurrence relations, demonstrating how to factor out constants, use mathematical induction, and derive the general form of the sequence. These techniques can be applied to a wide range of problems involving sequences and series, providing a strong foundation for further exploration into more complex mathematical concepts.