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Chapter 1: Problem 3

Prove the following statements for all positive integers: (a) \(1+3+5+\cdots+(2 n-1)=n^{2}\) (b) \(1^{2}+2^{2}+\cdots+n^{2}=n(n+1)(2 n+1) / 6\) (c) \(1^{3}+2^{3}+3^{3}+\cdots+n^{3}=[n(n+1) / 2]^{2}\)

Short Answer

Expert verified
In summary, we proved the following statements for all positive integers by using mathematical induction: (a) The sum of the first n odd numbers equals \(n^2\): \(1 + 3 + 5 + \cdots + (2n-1) = n^2\) (b) The sum of the squares of the first n natural numbers equals \(\frac{n(n+1)(2n+1)}{6}\): \(1^2 + 2^2 + \cdots + n^2 = \frac{n(n+1)(2n+1)}{6}\) (c) The sum of the cubes of the first n natural numbers equals \(\big(\frac{n(n+1)}{2}\big)^2\): \(1^3 + 2^3 + \cdots + n^3 = \big(\frac{n(n+1)}{2}\big)^2\)

Step by step solution

01

Base Case

For n=1: \(1 = 1^2\) Both sides are equal for n=1, so the base case holds true.
02

Induction Step

Assume the statement holds true for n=k: \(1 + 3 + 5 + \cdots + (2k - 1) = k^2\) Now, we need to show the statement holds true for n=k+1: \(1 + 3 + 5 + \cdots + (2k - 1) + (2(k+1) - 1) = (k+1)^2\)
03

Proving Inductive Hypothesis

We start by substituting the inductive hypothesis into the equation for n=k+1: \(k^2 + (2(k+1) - 1) = (k+1)^2\) Now, simplify and solve for k+1: \(k^2 + (2k+2-1) = (k^2+2k+1)\) \(k^2 + (2k+1) = (k^2+2k+1)\) Both sides are equal, so the statement holds true for n=k+1, completing the induction step and proving the statement for all positive integers. Statement (b) Prove: \(1^2 + 2^2 + \cdots + n^2 = \frac{n(n+1)(2n+1)}{6}\)
04

Base Case

For n=1: \(1^2 = \frac{1(1+1)(2(1)+1)}{6}\) \(1 = 1\) Both sides are equal for n=1, so the base case holds true.
05

Induction Step

Assume the statement holds true for n=k: \(1^2 + 2^2 + \cdots + k^2 = \frac{k(k+1)(2k+1)}{6}\) Now, we need to show the statement holds true for n=k+1: \(1^2 + 2^2 + \cdots + k^2 + (k+1)^2 = \frac{(k+1)((k+1)+1)(2(k+1)+1)}{6}\)
06

Proving Inductive Hypothesis

We start by substituting the inductive hypothesis into the equation for n=k+1: \(\frac{k(k+1)(2k+1)}{6} + (k+1)^2 = \frac{(k+1)(k+2)(2k+3)}{6}\) Now, simplify and solve for k+1: \((k+1)\big(\frac{k(2k+1)}{6} + (k+1)\big) = (k+1)\frac{(k+2)(2k+3)}{6}\) Divide both sides by (k+1), and we get: \(\frac{k(2k+1)}{6} + (k+1) = \frac{(k+2)(2k+3)}{6}\) Multiplying both sides by 6, we get: \(k(2k+1) + 6(k+1) = (k+2)(2k+3)\) Expanding and simplifying gives us: \(2k^2 + k + 6k + 6 = 2k^2 + 7k + 6\) Thus, both sides are equal, and the statement holds true for n=k+1, completing the induction step and proving the statement for all positive integers. Statement (c) Prove: \(1^3 + 2^3 + \cdots + n^3 = \big(\frac{n(n+1)}{2}\big)^2\)
07

Base Case

For n=1: \(1^3 = \big(\frac{1(1+1)}{2}\big)^2\) \(1 = 1\) Both sides are equal for n=1, so the base case holds true.
08

Induction Step

Assume the statement holds true for n=k: \(1^3 + 2^3 + \cdots + k^3 = \big(\frac{k(k+1)}{2}\big)^2\) Now, we need to show the statement holds true for n=k+1: \(1^3 + 2^3 + \cdots + k^3 + (k+1)^3 = \big(\frac{(k+1)((k+1)+1)}{2}\big)^2\)
09

Proving Inductive Hypothesis

We start by substituting the inductive hypothesis into the equation for n=k+1: \(\big(\frac{k(k+1)}{2}\big)^2 + (k+1)^3 = \big(\frac{(k+1)(k+2)}{2}\big)^2\) Now, simplify and solve for k+1: \((k+1)^2\big(\frac{k(k+1) + 2(k+1)}{4}\big) = (k+1)^2\frac{(k+2)^2}{4}\) Divide both sides by (k+1)^2, and we get: \(\frac{k(k+1) + 2(k+1)}{4} = \frac{(k+2)^2}{4}\) Multiplying both sides by 4, we get: \(k(k+1) + 2(k+1) = (k+2)^2\) Expanding and simplifying gives us: \(k^2 + k + 2k + 2 = k^2 + 4k + 4\) Thus, both sides are equal, and the statement holds true for n=k+1, completing the induction step and proving the statement for all positive integers.

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Key Concepts

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

Proof Techniques
Understanding proof techniques is crucial in mathematics to validate theorems and conjectures. There are several methods like direct proof, proof by contradiction, and proof by induction, which are used depending on the situation. Mathematical induction, the focus of our problem, is a proof technique often used to prove statements about integers. It consists of two steps: checking the base case and proving the so-called inductive step. This step-by-step approach ensures that if the proposition holds for an initial integer and if it also holds for an arbitrary integer implies it holds for the next integer, then the proposition is true for all integers. Induction's elegance lies in its simplicity; proving complex infinite series can break down into manageable chunks.
Series and Sequences
Series and sequences are fundamental concepts in mathematics, particularly in the field of algebra. A sequence is an ordered list of numbers following a particular rule, while a series is the sum of the elements of a sequence. Understanding the difference between the two and being able to work with their properties is crucial for solving problems in calculus, discrete mathematics, and various other fields. In our exercise, we deal with series of squares and cubes, asking us to sum these to arrive at a specific formula. These series have wide applications, from statistical calculations to computer algorithms.
Algebraic Equations
Algebraic equations are the bread and butter of algebra. These expressions set two quantities equal to each other and often contain one or more variables. In our series problems, algebraic equations emerge when we express the sum of a series as a function of n, the number of terms. Knowing how to manipulate these equations 鈥� which include operations like expansion, factorization, and simplification 鈥� is essential. Not only does this skill apply to series and sequences, but it also has wider applications in solving real-world problems where relationships between variables need to be established or discovered.
Inductive Step
The inductive step is a pillar of mathematical induction, as it propels the proof forward. After establishing the base case, we assume that our statement is true for some arbitrary positive integer k. This is known as the inductive hypothesis. The crux of the inductive step is to show that if the hypothesis is indeed true for k, it must also be true for k+1. The resolution of this step not only reinforces the hypothesis but ensures the theorem or formula at hand holds for all subsequent integers. This step requires both creative insights and algebraic skills, as we have seen in our exercise, where algebraic manipulations helped us prove the equations for both k and k+1.

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Most popular questions from this chapter

Let \(m, n\) be non-zero integers written in the form $$ m=p_{1}^{\prime \prime} \cdots p_{t}^{k} \quad \text { and } \quad n=p_{1}^{h} \cdots p_{r}^{b} . $$ where \(l_{v}, j\), are integers \(\geqq 0\) and \(p_{1} \ldots ., p_{r}\), are distinct prime numbers. (a) Show that the g.c.d. of \(m, n\) can be expressed as a product \(p_{1}^{k_{1}} \cdots p_{r}^{2,}\) where \(k_{1}, \ldots, k\), are integers \(\geq 0 .\) Express \(k_{v}\) in terms of \(i_{v}\) and \(f_{v}\) (b) Define the notion of least common multiple, and express the least common multiple of \(m, n\) as a product \(p_{1}^{k-\cdots} p_{f}^{k}\) with integers \(k_{v} \geqq 0 .\) Express \(k_{v}\) in terms of \(i_{v}\) and \(j_{v}\)

A positive integer is called palyndromic if its digits from left to right are the same as the digits from right to left. For instance, 242 and 15851 are palyndromic. The integers \(11,101,373,10301\) are palyndromic primes. Observe that except for 11, the others have an odd number of digits. (a) Is there a palyndromic prime with four digits? With an even number of digits fexcept for \(111 ?\) (b) Are there infinitely many palyndromic primes? (This is an unsolved problem in mathematics.)

Let \(n, d\) be positive integers and assume \(1

For all integers \(x, y\) and all primes \(p\) show that \((x+y)^{p} \equiv x^{p}+y^{p}(\bmod p)\).

(a) Prove that a positive integer is divisible by 3 if and only if the sum of its digits is divisible by 3 . (b) Prove that it is divisible by 9 if and only if the sum of its digits is divisible by \(9 .\) (c) Prove that it is divisible by 11 if and only if the alternating sum of its digits is divisible by 11. In other words. let the integer be $$ n=a_{k} a_{i-1} \cdots a_{0}=u_{0}+a_{1} 10+a_{2} 10^{2}+\cdots+a_{k} 10^{4}, \quad 0 \leqq a_{i} \leq 9. $$ Then \(n\) is divisible by 11 if and only if \(a_{0}-a_{1}+a_{2}-a_{3}+\cdots+(-1)^{4} a_{k}\) is divisible by 11
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