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Chapter 1: Problem 3

Use induction to prove the following statements for each \(n \in \mathbb{N}:\) (i) \(\sum_{i=1}^{n} i=\frac{n(n+1)}{2}\), (ii) \(\sum_{i=1}^{n} i^{2}=\frac{n(n+1)(2 n+1)}{6}\), (iii) \(\sum_{i=1}^{n} i^{3}=\frac{n^{2}(n+1)^{2}}{4}=\left(\sum_{i=1}^{n} i\right)^{2}\).

Short Answer

Expert verified
Using mathematical induction, we can prove the three statements: (i) \(\sum_{i=1}^{n} i=\frac{n(n+1)}{2}\) (ii) \(\sum_{i=1}^{n} i^{2}=\frac{n(n+1)(2 n+1)}{6}\) (iii) \(\sum_{i=1}^{n} i^{3}=\frac{n^{2}(n+1)^{2}}{4}=\left(\sum_{i=1}^{n} i\right)^{2}\) The process includes proving the base case (n=1), assuming it holds for n=k, and proving it for n=k+1, followed by concluding through the Principle of Mathematical Induction.

Step by step solution

01

Prove Statement (i) for n=1

For n=1, the left side of the equation is: \[ \sum_{i=1}^1 i = 1 \] For the right side of the equation, plug in n=1: \[\frac{1(1+1)}{2} = \frac{1 \cdot 2}{2} = 1\] Since the left side and right side are equal for n=1, the statement holds true for n=1.
02

Prove Statement (i) for n=k+1, assuming it holds for n=k

Assume the statement is true for n=k, that is \[\sum_{i=1}^k i = \frac{k(k+1)}{2}\] Now, let's prove it for n=k+1: \[\sum_{i=1}^{k+1} i = \left(\sum_{i=1}^k i\right) + (k+1)\] By assumption, \(\sum_{i=1}^k i = \frac{k(k+1)}{2}\). Thus, \[\left(\sum_{i=1}^k i\right) + (k+1) = \frac{k(k+1)}{2} + (k+1)\] Now we need to show that this expression is equal to \(\frac{(k+1)((k+1)+1)}{2}\). Simplify the expression: \[\frac{k(k+1)}{2} + (k+1) = \frac{k(k+1) + 2(k+1)}{2}\] Factor out (k+1), we get: \[\frac{(k+1)(k+2)}{2} = \frac{(k+1)((k+1)+1)}{2}\] Since the left side and right side are equal, the statement holds for n=k+1.
03

Conclude statement (i) by the Principle of Mathematical Induction

Since we proved statement (i) for n=1 and showed that it holds for n=k+1 given it holds for n=k, we can conclude by the Principle of Mathematical Induction that statement (i) is true for all integers n: \[\sum_{i=1}^{n} i=\frac{n(n+1)}{2}\]
04

Prove Statements (ii) and (iii) similarly

We can apply the same process to prove statements (ii) and (iii). For each statement, prove it for the base case (n=1), then assume it holds for n=k and prove it for n=k+1 using algebraic manipulation. Finally, conclude that each statement is true for all natural numbers n using the Principle of Mathematical Induction. Statement (ii): \[\sum_{i=1}^{n} i^{2}=\frac{n(n+1)(2 n+1)}{6}\] Statement (iii): \[\sum_{i=1}^{n} i^{3}=\frac{n^{2}(n+1)^{2}}{4}=\left(\sum_{i=1}^{n} i\right)^{2}\]

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Most popular questions from this chapter

Let \(I\) be an interval containing more than one point and \(f: I \rightarrow \mathbb{R}\) be any function. Given any distinct points \(x_{1}, x_{2}, x_{3} \in I\), define $$ \Psi\left(x_{1}, x_{2}, x_{3}\right)=\frac{\left(x_{3}-x_{2}\right) f\left(x_{1}\right)+\left(x_{1}-x_{3}\right) f\left(x_{2}\right)+\left(x_{2}-x_{1}\right) f\left(x_{3}\right)}{\left(x_{1}-x_{2}\right)\left(x_{2}-x_{3}\right)\left(x_{3}-x_{1}\right)} $$ (i) Show that \(\Psi\) is a symmetric function of \(x_{1}, x_{2}, x_{3}\), that is, show that \(\Psi\left(x_{1}, x_{2}, x_{3}\right)=\Psi\left(x_{2}, x_{1}, x_{3}\right)=\Psi\left(x_{3}, x_{2}, x_{1}\right)=\Psi\left(x_{1}, x_{3}, x_{2}\right)=\) \(\Psi\left(x_{2}, x_{3}, x_{1}\right)=\Psi\left(x_{3}, x_{1}, x_{2}\right)\) (ii) If \(\phi\) is as in Exercise 72 above, then show that $$ \Psi\left(x_{1}, x_{2}, x_{3}\right)=\frac{\phi\left(x_{1}, x_{3}\right)-\phi\left(x_{2}, x_{3}\right)}{x_{1}-x_{2}} \quad \text { for distinct } x_{1}, x_{2}, x_{3} \in I . $$ (iii) Show that \(f\) is convex on \(I\) if and only if \(\Psi\left(x_{1}, x_{2}, x_{3}\right) \geq 0\) for all distinct points \(x_{1}, x_{2}, x_{3} \in I\)

Show that \(n ! \leq 2^{-n}(n+1)^{n}\) for every \(n \in \mathbb{N}\), and that equality holds if and only if \(n=1\).

A set \(D\) is said to be countable if it is finite or if there is a bijective map from \(\mathbb{N}\) to \(D\). A set that is not countable is said to be uncountable. (i) Show that the set \(\\{0,1,2, \ldots\\}\) of all nonnegative integers is countable. (ii) Show that the set \(\\{1,3,5, \ldots\\}\) of all odd positive integers is countable. Also, the set \(\\{2,4,6, \ldots\\}\) of all even positive integers is countable. (iii) Show that the set \(\mathbb{Z}\) of all integers is countable.

Given any function \(f: \mathbb{R} \rightarrow \mathbb{R}\), prove the following: (i) If \(f(x y)=f(x)+f(y)\) for all \(x, y \in \mathbb{R}\), then \(f(x)=0\) for all \(x \in \mathbb{R}\). [Note: It is, however, possible that there are nonzero functions defined on subsets of \(\mathbb{R}\), such as \((0, \infty)\) that satisfy \(f(x y)=f(x)+f(y)\) for all \(x, y\) in the domain of \(f\). A prominent example of this is the logarithmic function, which will be discussed in Section 7.1.] (ii) If \(f(x y)=f(x) f(y)\) for all \(x, y \in \mathbb{R}\), then either \(f(x)=0\) for all \(x \in \mathbb{R}\), or \(f(x)=1\) for all \(x \in \mathbb{R}\), or \(f(0)=0\) and \(f(1)=1\). Further, \(f\) is either an even function or an odd function. Give examples of even as well as odd functions \(f: \mathbb{R} \rightarrow \mathbb{R}\) satisfying \(f(x y)=f(x) f(y)\) for all \(x, y \in \mathbb{R}\)

Let \(S\) be a nonempty subset of \(\mathbb{R}\). If \(S\) is bounded above, then show that the set \(U_{S}=\\{\alpha \in \mathbb{R}: \alpha\) is an upper bound of \(S\\}\) is bounded below, \(\min U_{S}\) exists, and \(\sup S=\min U_{S} .\) Likewise, if \(S\) is bounded below, then show that the set \(L_{S}=\\{\beta \in \mathbb{R}: \beta\) is a lower bound of \(S\\}\) is bounded above, \(\max L_{S}\) exists, and inf \(S=\max L_{S}\)
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