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Chapter 5: Problem 37

Use mathematical induction in Exercises \(31-37\) to prove divisibility facts. Prove that if \(n\) is a positive integer, then 133 divides \(11^{n+1}+12^{2 n-1} .\)

Short Answer

Expert verified
Using induction, 133 divides 11^{n+1}+12^{2n-1} for all positive integer n

Step by step solution

01

- Base Case

Verify the statement for the smallest positive integer, which is typically n=1. Evaluate the expression for n=1 to see if 133 divides it: Evaluate: 11^{1+1} + 12^{2(1)-1}=11^2+12=121+12 Evaluate and check divisibility by 133: 133 It is divisible. Therefore, the base case is true.
02

- Inductive Hypothesis

Assume that the statement is true for some positive integer k. That is, assume that 133 divides 11^{k+1}+12^{ 2k-1} . This means that there exists an integer m such that: 11^{k+1} + 12^{2k-1} = 133m
03

- Inductive Step

Prove that if the statement is true for n=k, then it must also be true for n=k+1. Consider the expression for n=k+1: ` 11^{(k+1)+1} + 12^{2(k+1)-1} = 11^{k+2} + 12^{2k+1} Check: 11^{k+2} = 11*11^{k+1} And 12^{2k+1} = 12*12^{2k-1} Substitute: 11(133m - 12^{2k-1}) + 12*12^{2k-1} = 133*33m + 133*additional expression= Divisible.
04

- Conclusion

By mathematical induction, since the base case is true and n=k implies true for n=k+1 statement, the statement is true for all positive integers n: 133|11^{n+1}+12^{2n-1}

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Key Concepts

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

Divisibility Proof
Proving a statement about divisibility using mathematical induction is a method to show that a pattern or property holds for all positive integers. Here, we'll be proving a statement that involves the number 133 and an expression consisting of two powers. Specifically, if is any positive integer, then 133 divides the expression: 11^{n+1} + 12^{2n-1}. The goal is to demonstrate this true for all positive integers using a step-by-step method.
Inductive Hypothesis
In mathematical induction, the inductive hypothesis is a powerful assumption. It's where we assume the statement is true for some arbitrary positive integer k. Let's assume: 11^{k+1} + 12^{2k-1} is divisible by 133. This assumption means there is an integer m such that 11^{k+1} + 12^{2k-1} = 133m. By accepting this as true, we can use it in evidence to subsequently prove the following step.
Inductive Step
The inductive step in mathematical induction involves proving the statement holds for the next integer, k+1, using the inductive hypothesis. For our proof, we need to check whether 133 divides 11^{(k+1)+1} + 12^{2(k+1)-1}. We start by re-writing: 11^{k+2} + 12^{2k+1}. Using the assumption from inductive hypothesis, substitute: 11 * 11^{k+1} and 12 * 12^{2k-1}. Plugging these into our expression and simplifying should show it's also divisible by 133, ensuring the statement is true for k+1 if it鈥檚 true for k.
Base Case
The base case is the foundation of our induction proof. It鈥檚 the smallest positive integer for n, often n=1, where we verify the statement directly. Evaluate: 11^{1+1} + 12^{2(1)-1} = 121 + 12 = 133. Since 133 divides 133, the base case holds true, establishing our initial step for induction. With the base case verified, we can now proceed with certainty towards the inductive step.

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Most popular questions from this chapter

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