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Chapter 8: Problem 14

Show that the sequence \(0,1,5,19, \ldots, 3^{n}-2^{n}, \ldots\), defined for \(n \geq 0\), satisfies the recurrence relation $$ d_{k}=5 d_{k-1}-6 d_{k-2} \quad \text { for all integers } k \geq 2 \text {. } $$

Short Answer

Expert verified
In conclusion, we have used induction to prove that the sequence \(0, 1, 5, 19, \ldots, 3^n - 2^n, \ldots\) for \(n \geq 0\) satisfies the recurrence relation \(d_k = 5d_{k-1} - 6d_{k-2}\) for all integers \(k \geq 2\). The base case holds true for \(k=2\), and the inductive step was shown to be true for \(k + 1\) given that it's true for \(k\).

Step by step solution

01

Understanding the sequence and the recurrence relation

We have the sequence given by the formula \(d_{n} = 3^n - 2^n\) for all integers \(n \geq 0\). The recurrence relation is given by \(d_{k} = 5d_{k-1} - 6d_{k-2}\) for all integers \(k \geq 2\). To show that the sequence satisfies the recurrence relation, we need to demonstrate that for all \(k \geq 2\), we can use the relation to find the next term of the sequence given the previous two terms.
02

Base case for induction

Let's prove that the statement is true for the base case when \(k=2\). According to the sequence, \(d_{0}=3^0-2^0 =0\), \(d_1= 3^1-2^1=1\), and \(d_2= 3^2-2^2=5\). Now let's plug these values into the recurrence relation: \(d_{2} = 5d_{1} - 6d_{0} = 5(1) - 6(0) = 5\) The base case holds true since the value of \(d_2\) from the sequence and the value using the recurrence relation are both equal to 5.
03

Inductive hypothesis

Now assume that the statement holds true for some integer \(k \geq 2\). That is, \(d_{k} = 5d_{k-1} - 6d_{k-2}\) We need to show that the statement is true for the next integer \(k + 1\). This means we need to demonstrate that \(d_{k+1} = 5d_{k} - 6d_{k-1}\)
04

Inductive step

To demonstrate the inductive step, let's express \(d_{k+1}\), \(d_k\), and \(d_{k-1}\) in terms of \(3^n\) and \(2^n\). We know that: \(d_{k+1} = 3^{k+1} - 2^{k+1}\) \(d_k = 3^k - 2^k\) \(d_{k-1} = 3^{k-1} - 2^{k-1}\) Now let's plug these expressions into the equation we want to prove: \(5(3^k - 2^k) - 6(3^{k-1} - 2^{k-1})\) By simplifying, we get: \(5\cdot 3^k - 5\cdot 2^k - 6\cdot 3^{k-1} + 6\cdot 2^{k-1}\) \(= 3^{k-1}(5\cdot 3) - 2^{k-1}(5\cdot 2) + 6\cdot 2^{k-1} - 6\cdot 3^{k-1}\) \(= 3^{k-1}(3^2 - 6) + 2^{k-1}(6-2^2)\) \(= 3^{k-1} \cdot 3^2 - 2^{k-1} \cdot 2^2 = 3^{k+1} - 2^{k+1}=d_{k+1}\) Hence, by induction, the recurrence relation \(d_{k} = 5d_{k-1} - 6 d_{k-2}\) holds for all integers \(k \geq 2\).
05

Conclusion

In conclusion, we have shown through a proof by induction that the given sequence \(0, 1, 5, 19, \ldots, 3^n - 2^n, \ldots\) for \(n \geq 0\) satisfies the recurrence relation \(d_k = 5d_{k-1} - 6d_{k-2}\) for all integers \(k \geq 2\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Mathematical Induction
Mathematical induction is a powerful proof technique used for establishing the truth of an infinite number of statements. It works similarly to dominoes falling 鈥� if you can knock the first one and each subsequent domino can knock the next, all will fall. That's how induction works, too!
To apply mathematical induction, we use the following steps:
  • Base Case: We start by proving that the statement holds for the initial position, typically when the variable equals zero or one.
  • Inductive Hypothesis: Assume that it holds for an arbitrary positive integer k.
  • Inductive Step: Prove that if the statement holds for k, then it must also be true for k+1.
By ensuring the base case and showing the sequence of implications, we confirm the property for all natural numbers beyond the base case. The exercise demonstrated this method by proving the sequence follows the given recurrence relation.
Sequences
A sequence is an ordered list of numbers following a particular pattern. Each number in a sequence is called a term. For mathematicians and computer scientists, finding the rules that govern sequences lets us predict terms further down the list without actually listing all prior terms.
In the given exercise, we look at a sequence defined by the formula \(d_n = 3^n -2^n\). Each term depends on the natural number n. It is crucial to note how sequences can also be described using recurrence relations, which are formulas that express each term in terms to one or more of the previous terms.
The sequence here is seen to satisfy a recurrence relation. By determining such relationships, the method provides insights into how each part of the sequence fits together systematically, making calculations of future terms easier.
Proofs in Discrete Mathematics
Proofs are the backbone of discrete mathematics, providing the means to verify the truth of mathematical statements. In discrete mathematics, proofs can be presented in various forms, such as direct proofs, proof by contradiction, or the mathematical induction method.
For this exercise, proof through induction was utilized to show that the sequence adheres to the recurrence relation \(d_k = 5d_{k-1} - 6d_{k-2}\). This started with the base case to establish the foundation and progressed through the inductive step to build on that foundation and confirm the entire structure holds true.
The elegance of proofs in discrete mathematics lies in their clarity and logical structure, offering clear stepping stones from assumptions to conclusions. They provide rigorous justification that mathematicians and students rely on to make confident assertions about mathematical objects and their properties.

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