/*! 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 18 Prove that \(2+5+8+\cdots+(3 n-1... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 18

Prove that \(2+5+8+\cdots+(3 n-1)=\frac{1}{2} n(3 n+1)\) for all \(n \in \mathbb{N}\)

Short Answer

Expert verified
Using the method of mathematical induction, we first verify the base case for n = 1: \[\frac{1}{2}(1)(4) = 2\] Next, we assume the formula is true for n = k and add the next term in the sequence to show that the formula is true for n = k + 1: \[\frac{1}{2} k(3k+1) + (3k+2) \] Simplifying this expression, we get: \[\frac{1}{2}(3k^2 + 7k + 4)\] which is equal to the formula for n = k + 1: \[\frac{1}{2} (k+1)(3k+4)\] Hence, we have proven that the formula \[2+5+8+\cdots+(3n-1)=\frac{1}{2}n(3n+1)\] holds true for all natural numbers n.

Step by step solution

01

The base case

For our induction method to work, we need to show that the formula is true for n = 1. In this case, the sum is simply the first term of the sequence, which is 2. So, let's plug n = 1 into the formula and see if it holds: \[\frac{1}{2} (1)(3(1)+1) = \frac{1}{2} (4) = 2\] Since the left-hand side of the equation equals the right-hand side of the equation, the formula holds for n = 1.
02

The inductive step

Now let's assume that the formula is true for n = k: \[2+5+8+\cdots+(3k-1)=\frac{1}{2} k(3 k+1)\] Our goal is to prove that the formula holds for n = k + 1. To do this, we need to add the next term in the sequence, which is 3(k + 1) - 1. Let's see what happens when we do that: \[2+5+8+\cdots+(3k-1) + (3(k+1)-1) = \frac{1}{2} k(3 k+1) + (3(k+1)-1)\] We want to show that this is equal to the formula for n = k + 1: \[\frac{1}{2} (k+1)(3(k+1) +1)\]
03

Simplifying the equation

Next, we will simplify the equation from Step 2: \[\frac{1}{2} k(3 k+1) + (3(k+1)-1) = \frac{1}{2} k(3 k+1) + (3k+2)\] Now, we want to show that this is equal to the formula for n = k + 1: \[\frac{1}{2} (k+1)(3(k+1) +1) = \frac{1}{2} (k+1)(3k + 4)\] Multiplying out the right side yields: \[\frac{1}{2} (k+1)(3k + 4) = \frac{1}{2} (3k^2 + 7k + 4)\]
04

Comparing the equation from Step 2 with the simplified equation

Now, we will process the left-hand side of the equation in Step 2. \[\frac{1}{2} k(3 k+1) + (3k+2) = \frac{1}{2}(3k^2 + k + 6k + 4)\] Simplifying this expression: \[\frac{1}{2}(3k^2 + 7k + 4)\] Comparing this expression with the right-hand side of the expression from Step 3, we can see that they are indeed equal.
05

Completing the induction

Since we have shown that the formula is true for the base case (n = 1), and we have shown that if the formula holds for n = k, it also holds for n = k + 1, we can conclude that the formula: \[2+5+8+\cdots+(3 n-1)=\frac{1}{2} n(3 n+1)\] holds true for all natural numbers 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 91影视!

Key Concepts

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

Proof Techniques
Understanding proof techniques is crucial for solving mathematical problems and asserting the validity of various propositions within the realm of mathematics. One effective approach is mathematical induction, which operates on the principle of establishing the truth of an infinite sequence of propositions. The technique involves two pivotal steps to demonstrate the validity of a statement for all natural numbers.

Initially, a base case is considered; this is the simplest case, often where the natural number is 1. If the statement holds true for the base case, an inductive step follows. In the inductive step, the truth of the statement for a particular natural number, say k, is assumed, and then it is used to show that the statement must also be true for the next number in the sequence, k + 1.

If both the base case and the inductive step are verified, induction allows us to infer that the statement is true for all natural numbers. This leap from one instance to all instances distinguishes mathematical induction from direct proof and provides a powerful tool for handling propositions about infinite sequences.
Arithmetic Series
Arithmetic series plays a significant role in understanding patterns within sequences of numbers. It is defined as the sum of a sequence where each term is generated by adding a constant to the previous term. In the context of the exercise, we are dealing with an arithmetic series where the common difference is three: 2, 5, 8, and so on.

The general formula for the sum of the first n terms of an arithmetic series is given by \( S_n = \frac{n}{2}(a_1 + a_n) \), where \( a_1 \) is the first term and \( a_n \) is the nth term. This formula arises from the fact that the sum of evenly spaced numbers can be grouped into pairs that have the same total. The series in our exercise can also be expressed using a general formula, which we aim to prove using mathematical induction.

The concept of arithmetic series not only provides a method to sum large sets of numbers rapidly but also helps in understanding other mathematical concepts such as infinite series and convergence, which are vital in various branches of mathematics and applied sciences.
Natural Numbers
Natural numbers are the set of positive integers that we use for counting and ordering. In mathematics, they are denoted by \( \mathbb{N} \) and typically start from 1, extending indefinitely to include 2, 3, 4, and so forth. An important property of natural numbers is that they are infinite; there is no largest natural number.

When working with proofs involving natural numbers, we often use mathematical induction because it leverages this endless nature of natural numbers. The exercise provided exemplifies the applicability of induction for statements where natural numbers are involved. As it operates under the assumption that if something is true for one natural number and its truth leads to the truth of the next, it fits perfectly with the sequential and infinitely stretching characteristics of natural numbers.

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

Prove that \((2)(6)(10)(14) \cdots(4 n-2)=\frac{(2 n) !}{n !}\), for all \(n \in \mathbb{N}\).

Conjecture a formula for the sum \(5+9+13+\cdots+(4 n+1)\), and prove your conjecture using mathematical induction. t?

Prove that \((\cos x+i \sin x)^{n}=\cos (n x)+i \sin (n x)\), for all \(n \in \mathbb{N}\), where \(i=\) \(\sqrt{-1}\). You may use the identities \(\cos (a+b)=\cos a \cos b-\sin a \sin b\) and \(\sin (a+b)=\sin a \cos b+\cos a \sin b\).

Prove that \(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\cdots+\frac{1}{4 n^{2}-1}=\frac{n}{2 n+1}\), for all \(n \in \mathbb{N} .\)

Prove that \(9^{n}-4^{n}\) is a multiple of 5 for all \(n \in \mathbb{N}\)
See all solutions

Recommended explanations on Math Textbooks

Mechanics Maths

Read Explanation

Geometry

Read Explanation

Applied Mathematics

Read Explanation

Logic and Functions

Read Explanation

Discrete Mathematics

Read Explanation

Theoretical and Mathematical Physics

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.