/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 12 If \(n \in \mathbf{Z}^{+}\), pro... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

If \(n \in \mathbf{Z}^{+}\), prove that 57 divides \(7^{n+2}+8^{2 n+1}\).

Short Answer

Expert verified
By the principle of mathematical induction, the statement that 57 divides \(7^{n+2}+8^{2n+1}\) is true for any positive integer \(n\).

Step by step solution

01

Base Case

Evaluate when \(n = 1\). This gives \(7^{1+2}+8^{2*1+1}=7^{3}+8^{3}=504\). 57 does divide 504 (504/57 = 8.84), confirming our base case.
02

Inductive Step

Assume the statement is true for some positive integer \(k\). This means that 57 can divide \(7^{k+2}+8^{2k+1}\). Now, we need to show that the statement holds for \(k + 1\). That is, we need to prove that 57 can divide \(7^{k+3}+8^{2k+3}\).
03

Prove k + 1

This is the most complicated step. Express \(7^{k+3}+8^{2k+3}\) as \(7*7^{k+2}+8^{2}*8^{2k+1}\). Using the inductive assumption, the term \(7^{k+2}\) can be written as \(57m-8^{2k+1}\) for some integer \(m\), and the term \(8^{2k+1}\) as \(57n-7^{k+2}\) for some integer \(n\). We get \(7^{k+3}+8^{2k+3}\) equals to \(7(57m-8^{2k+1})+2*8(57n-7^{k+2})\). This simplifies to \(57(7m-14n)+8^{2k+3}-14*7^{k+2}\). Here, \(8^{2k+3}-14*7^{k+2}\) equals to \(2*8^{2k+1}-14*7^{k+2}\) which can be written as \(2*(57x-7^{k+2})-14*7^{k+2}\) for some integer \(x\). After simplifying, this expression can be written as \(57*\text{integer}\), proving the statement for \(k+1\) holds if it holds for \(k\). Thus, by the principle of mathematical induction, we can say that the statement is true for any positive integer \(n\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Proof by Induction
Proof by induction is a powerful mathematical technique used to prove that a proposition holds true for all natural numbers. It works on the principle of domino effect; if you can knock down the first domino (the base case) and show that knocking down one domino will always knock down the next (the inductive step), then all the dominos will fall. In the context of our problem where we want to prove that 57 divides \(7^{n+2}+8^{2n+1}\) for all positive integers \(n\), we start with the base case of \(n=1\).

We confirmed that 57 indeed divides \(7^{1+2}+8^{2*1+1}\). After establishing the base case, we move to the inductive step. This involves assuming that our proposition holds true for some positive integer \(k\), which implies that 57 divides \(7^{k+2}+8^{2k+1}\). We then show that if this is true, it must also be true for \(k+1\) by performing algebraic manipulations and utilizing the inductive hypothesis. This step confirms that if the proposition is true for \(k\), it is also true for \(k+1\), thereby proving by induction that the proposition is true for all positive integers.

To enhance understanding, it's crucial to discuss some of the common pitfalls students may encounter during induction proofs. It's not uncommon to see mistakes in the algebraic manipulations or forgetting to explicitly state the inductive hypothesis. Careful step-by-step expansion and simplification of terms are essential to avoid these errors and ensure the proof is valid.
Divisibility
Divisibility in mathematics refers to the ability of one number to be divided by another without leaving a remainder. In the task of proving divisibility through induction, we aim to show that one number (in this case, 57) divides a complex expression without a remainder for all values of \(n\).

In the given exercise, we are given two expressions, \(7^{n+2}\) and \(8^{2n+1}\), and the goal is to show that their sum is divisible by 57 for any positive integer \(n\). This demonstration involves clever algebraic manipulation where the expressions are rewritten using the previously established inductive assumption. At each step of the solution, we focus on maintaining the form that reveals divisibility, for example, expressing part of the complex expression in the form of \(57m\), where \(m\) is an integer, reinforces the notion that the entire expression maintains divisibility by 57.

To facilitate understanding, keeping an eye out for the structure similar to \(57\times integer\) within the calculations is a key strategy. Students should be encouraged to understand rather than memorize steps, as this approach promotes a deeper comprehension of how divisibility interacts with algebraic expressions in proofs.
Exponentiation
Exponentiation is the mathematical operation involving two numbers, the base and the exponent, where the base is raised to the power of the exponent. In our exercise, we see exponentiation in \(7^{n+2}\) and \(8^{2n+1}\). Understanding how to work with exponential expressions is critical, especially when they are part of a larger mathematical proof, like induction.

When performing proof by induction that involves exponentiation, special attention must be given to how these expressions can be manipulated while maintaining the integrity of the proof. For example, when examining the transformation of \(7^{k+3}\) into \(7 \cdot 7^{k+2}\), it's a utilization of the exponentiation rule \(a^{m+n} = a^m \cdot a^n\).

Moreover, in inductive proofs involving exponentiated terms, it is also common to factor out powers to simplify the expressions in a way that reveals patterns conducive to the induction process. An essential improvement tip is to remind students to always revert to the basic properties of exponentiation when simplifying terms during these proofs. This includes leveraging the distributive property to break down expressions and understand the relationship between the resulting terms, which is vital in proofs involving variable exponents.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Let \(a, b, c \in \mathbf{Z}^{+}\)with \(\operatorname{gcd}(a, b)=1\). If \(a \mid c\) and \(b \mid c\), prove that \(a b \mid c\). Does the result hold if \(\operatorname{gcd}(a, b) \neq 1\) ?

Write each of the following (base-10) integers in base 2 and base 16 . a) 22 b) 527 c) 1234 d) 6923

Consider the permutations of \(1,2,3,4\). The permutation 1432, for instance, is said to have one ascent - namely, 14 (since \(1<4)\). This same permutation also has two descents namely, 43 (since \(4>3)\) and 32 (since \(3>2\) ). The permutation 1423 , on the other hand, has two ascents, at 14 and 23 - and the one descent 42 . a) How many permutations of \(1,2,3\) have \(k\) ascents, for \(k=0,1,2 ?\) b) How many permutations of \(1,2,3,4\) have \(k\) ascents, for \(k=0,1,2,3 ?\) d) Suppose a permutation of \(1,2,3, \ldots, m\) has \(k\) ascents, for \(0 \leq k \leq m-1\). How many descents does the permutation have? e) Consider the permutation \(p=12436587\). This permutation of \(1,2,3, \ldots, 8\) has four ascents. In how many of the nine locations (at the start, end, or between two numbers) in \(p\) can we place 9 so that the result is a permutation of \(1,2,3, \ldots, 8,9\) with (i) four ascents; (ii) five ascents? f) Let \(\pi_{m, k}\) denote the number of permutations of \(1,2,3\), \(\ldots, m\) with \(k\) ascents. Note how \(\pi_{4,2}=11=2(4)+\) \(3(1)=(4-2) \pi_{3,1}+(2+1) \pi_{3,2} .\) How is \(\pi_{m, k}\) related to \(\pi_{m-1, k-1}\) and \(\pi_{m-1, k}\) ? c) If a permutation of \(1,2,3,4,5,6,7\) has four ascents, how many descents does it have?

a) How many positive divisors are there for $$ n=2^{14} 3^{9} 5^{8} 7^{10} 11^{3} 13^{5} 37^{10} ? $$ b) For the divisors in part (a), how many are i) divisible by \(2^{3} 3^{4} 5^{7} 11^{2} 37^{2} ?\) ii) divisible by \(1,166,400,000\) ? iii) perfect squares? iv) perfect squares that are divisible by \(2^{2} 3^{4} 5^{2} 11^{2} ?\) v) perfect cubes? vi) perfect cubes that are multiples of \(2^{10} 3^{9} 5^{2} 7^{5} 11^{2} 13^{2} 37^{2} ?\) vii) perfect squares and perfect cubes?

a) Let \(a \in \mathbf{Z}^{+}\). Prove or disprove: (i) If \(10 \mid a^{2}\), then \(10 \mid a\); and (ii) If \(4 \mid a^{2}\), then \(4 \mid a\). b) Generalize the true result(s) in part (a).

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.