/*! 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 3 Use mathematical induction to pr... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 3

Use mathematical induction to prove each of the following: (a) For each natural number \(n, 2+5+8+\cdots+(3 n-1)=\frac{n(3 n+1)}{2}\). (b) For each natural number \(n, 1+5+9+\cdots+(4 n-3)=n(2 n-1)\). (c) For each natural number \(n, 1^{3}+2^{3}+3^{3}+\cdots+n^{3}=\left[\frac{n(n+1)}{2}\right]^{2}\).

Short Answer

Expert verified
We have proved all three statements using mathematical induction by demonstrating their validity for n = 1 (base case), and then assuming that each formula holds for n = k, we showed it also holds for n=k+1 (inductive step). Thus, the given formulas are valid for all natural numbers n.

Step by step solution

01

Base Case

When n = 1, the formula becomes: \[2 = \frac{1\times(3\times1+1)}{2}\] Simplifying the right side of the equation: \[2 = \frac{1\times4}{2} = 2\] The formula holds for n=1.
02

Inductive Step

Now let's assume that the formula holds for n=k (Inductive hypothesis): \[2+5+8+\cdots+(3k-1)=\frac{k(3k+1)}{2}\] We want to show that this formula also holds for n=k+1: \[2+5+8+\cdots+(3k-1) + (3(k+1)-1) = \frac{(k+1)(3(k+1)+1)}{2}\] Using the inductive hypothesis, we can rewrite the left side: \[\frac{k(3k+1)}{2} + (3(k+1)-1) = \frac{(k+1)(3(k+1)+1)}{2}\] Now, we have to prove that the left side is equal to the right side: \[\frac{k(3k+1)+2(3k+2)}{2} = \frac{(k+1)(3k+4)}{2}\] \[k(3k+1) + 6k+4 = (k+1)(3k+4)\] Expanding and simplifying, we get: \[3k^2+k+6k+4 = 3k^2+4k+3k+4\] Which is a true statement. Therefore, the formula holds true for n=k+1, completing the induction proof for (a). ##Proof for (b)## Now we will prove the formula in (b) using mathematical induction.
03

Base Case

When n = 1, the formula becomes: \[1 = 1\times(2\times1-1)\] Simplifying the right side of the equation: \[1 = 1\times1 = 1\] The formula holds for n=1.
04

Inductive Step

Now let's assume that the formula holds for n=k (Inductive hypothesis): \[1+5+9+\cdots+(4k-3)=k(2k-1)\] We want to show that this formula also holds for n=k+1: \[1+5+9+\cdots+(4k-3) + (4(k+1)-3) = (k+1)(2(k+1)-1)\] Using the inductive hypothesis, we can rewrite the left side: \[k(2k-1) + (4k+1) = (k+1)(2k+1)\] Now, we have to prove that the left side is equal to the right side: \[2k^2-k+4k+1 = 2k^2+2k+k+1\] Combining like terms, we get: \[2k^2+3k+1 = 2k^2+3k+1\] Which is a true statement. Therefore, the formula holds true for n=k+1, completing the induction proof for (b). ##Proof for (c)## Now we will prove the formula in (c) using mathematical induction.
05

Base Case

When n = 1, the formula becomes: \[1^3 = \left[\frac{1\times(1+1)}{2}\right]^2\] Simplifying the right side of the equation: \[1 = \left[\frac{2}{2}\right]^2 = 1\] The formula holds for n=1.
06

Inductive Step

Now let's assume that the formula holds for n=k (Inductive hypothesis): \[1^3 + 2^3 + 3^3 +\cdots+ k^3 = \left[\frac{k(k+1)}{2}\right]^2\] We want to show that this formula also holds for n=k+1: \[1^3 + 2^3 + 3^3 +\cdots+ k^3 + (k+1)^3 = \left[\frac{(k+1)(k+2)}{2}\right]^2\] Using the inductive hypothesis, we can rewrite the left side: \[\left[\frac{k(k+1)}{2}\right]^2 + (k+1)^3 = \left[\frac{(k+1)(k+2)}{2}\right]^2\] Now, we have to prove that the left side is equal to the right side: \[\frac{k^2(k+1)^2 + 4(k+1)^3}{4} = \frac{(k+1)^2(k+2)^2}{4}\] We can cancel out the common terms: \[k^2(k+1) + 4(k+1)^2 = (k+1)(k+2)^2\] Factor out (k+1) from the left side: \[(k+1)(k^2+4(k+1)) = (k+1)(k+2)^2\] Dividing both sides by (k+1), we get: \[k^2+4(k+1) = (k+2)^2\] Expanding and simplifying, we get: \[k^2+4k+4 = k^2+4k+4\] Which is a true statement. Therefore, the formula holds true for n=k+1, completing the induction proof for (c).

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 91影视!

Key Concepts

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

Base Case
The base case is the starting point of a mathematical induction proof. It involves verifying that the statement or formula is true for the initial case, typically when the natural number is equal to 1. For instance, in the example given for part (a) of the exercise, the base case is when \(n = 1\). Here, you calculate both sides of the equation independently: the sum is \(2\), and the formula given is \(\frac{1(3 \times 1 + 1)}{2}\). Simplifying the right side gives \(2\), confirming that the base case holds. This initial verification is crucial, as it establishes the foundation from which the induction process can begin.
Inductive Hypothesis
The inductive hypothesis is a key step in mathematical induction. After proving the base case, you assume that the statement is true for a specific natural number, say \(n = k\). This assumption allows you to use the formula or statement as a given for this particular instance. In the example of part (a), the inductive hypothesis assumes the formula holds true for \(n = k\):
  • \(2+5+8+\cdots+(3k-1)=\frac{k(3k+1)}{2}\)
This assumption sets the stage for verifying the formula for the next natural number, \(k + 1\). The inductive hypothesis provides a bridge from one instance to the next in the sequence of natural numbers.
Inductive Step
The inductive step is where the heart of the mathematical induction proof lies. It involves showing that if the inductive hypothesis holds for some natural number \(n = k\), it must also hold for \(n = k+1\). This step requires you to use the assumed statement for \(k\) to prove it true for \(k+1\). In our example, for part (a), you would need to demonstrate:
  • \(2+5+8+\cdots+(3k-1) + (3(k+1)-1) = \frac{(k+1)(3(k+1)+1)}{2}\)
By substituting the inductive hypothesis into the equation, you combine and simplify both sides to show they are equal. If this is successful, the formula stands confirmed for \(k+1\). It's this repetition across all natural numbers that proves the statement universally.
Natural Numbers
Natural numbers form the essence of mathematical induction. They are the set of positive integers beginning from 1 and extending infinitely upwards (1, 2, 3, 4, ...). In proofs by mathematical induction, statements are often made about these numbers, as they are fundamental to number theory and discrete mathematics. In the context of our exercise, sorting through each natural number sequentially helps show that the formula holds across the board. Understanding natural numbers and their properties is crucial, as it ensures that the induction process can be applied effectively and comprehensively to derive true mathematical conclusions. These numbers serve as the conceptual framework within which the induction process unfolds.

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

For the sequence \(a_{1}, a_{2}, \ldots, a_{n}, \ldots,\) assume that \(a_{1}=1, a_{2}=5,\) and that for each \(n \in \mathbb{N}, a_{n+1}=a_{n}+2 a_{n-1}\). Prove that for each natural number \(n\), \(a_{n}=2^{n}+(-1)^{n}\)

(a) Prove that if \(n \in \mathbb{N},\) then there exists an odd natural number \(m\) and a nonnegative integer \(k\) such that \(n=2^{k} m\). (b) For each \(n \in \mathbb{N}\), prove that there is only one way to write \(n\) in the form described in Part (a). To do this, assume that \(n=2^{k} m\) and \(n=2^{q} p\) where \(m\) and \(p\) are odd natural numbers and \(k\) and \(q\) are nonnegative integers. Then prove that \(k=q\) and \(m=p\).

Evaluation of proofs See the instructions for Exercise (19) on page 100 from Section 3.1 . (a) Let \(f_{n}\) be the \(n^{\text {th }}\) Fibonacci number, and let \(\alpha\) be the positive solution of the equation \(x^{2}=x+1 .\) So \(\alpha=\frac{1+\sqrt{5}}{2}\). For each natural number \(n, f_{n} \leq \alpha^{n-1}\) Proof. We will use a proof by mathematical induction. For each natural number \(n,\) we let \(P(n)\) be, " \(f_{n} \leq \alpha^{n-1}\) " We first note that \(P(1)\) is true since \(f_{1}=1\) and \(\alpha^{0}=1\). We also notice that \(P(2)\) is true since \(f_{2}=1\) and, hence, \(f_{2} \leq \alpha^{1}\). We now let \(k\) be a natural number with \(k \geq 2\) and assume that \(P(1)\), \(P(2), \ldots, P(k)\) are all true. We now need to prove that \(P(k+1)\) is true or that \(f_{k+1} \leq \alpha^{k}\). Since \(P(k-1)\) and \(P(k)\) are true, we know that \(f_{k-1} \leq \alpha^{k-2}\) and \(f_{k} \leq \alpha^{k-1}\). Therefore, $$ \begin{aligned} f_{k+1} &=f_{k}+f_{k-1} \\ f_{k+1} & \leq \alpha^{k-1}+\alpha^{k-2} \\ f_{k+1} & \leq \alpha^{k-2}(\alpha+1) . \end{aligned} $$ We now use the fact that \(\alpha+1=\alpha^{2}\) and the preceding inequality to obtain $$ \begin{array}{l} f_{k+1} \leq \alpha^{k-2} \alpha^{2} \\ f_{k+1} \leq \alpha^{k} \end{array} $$ This proves that if \(P(1), P(2), \ldots, P(k)\) are true, then \(P(k+1)\) is true. Hence, by the Second Principle of Mathematical Induction, we conclude that or each natural number \(n, f_{n} \leq \alpha^{n-1}\).

Instead of using induction, we can sometimes use previously proven results about a summation to obtain results about a different summation. (a) Use the result in Progress Check 4.3 to prove the following proposition: For each natural number \(n, 3+6+9+\cdots+3 n=\frac{3 n(n+1)}{2}\) (b) Subtract \(n\) from each side of the equation in Part (a). On the left side of this equation, explain why this can be done by subtracting 1 from each term in the summation. (c) Algebraically simplify the right side of the equation in Part (b) to obtain a formula for the sum \(2+5+8+\cdots+(3 n-1)\). Compare this to Exercise (3a).

Can each natural number greater than or equal to 4 be written as the sum of at least two natural numbers, each of which is a 2 or a 3 ? Justify your conclusion. For example, \(7=2+2+3,\) and \(17=2+2+2+2+3+3+3\).
See all solutions

Recommended explanations on Math Textbooks

Applied Mathematics

Read Explanation

Logic and Functions

Read Explanation

Calculus

Read Explanation

Geometry

Read Explanation

Pure Maths

Read Explanation

Probability and Statistics

Read Explanation
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.