Explain. In this case, we use \(j\) for the index for the summation, and the notation \(\sum_{j=1}^n j^2\) tells us to add all the values of \(j^2\) for \(j\) from 1 to \(n\), inclusive. For every natural number \(n\), \(5^n \equiv 1\) (mod 4). For another example, for each natural number \(n\), we now let \(Q(n)\) be the following open sentence: \[1^2 + 2^2 + ... + n^2 = \dfrac{n(n + 1)(2n + 1)}{6}.\]. We resolve this by making Statement (1) an axiom for the natural numbers so that this becomes one of the defining characteristics of the natural numbers. \ \ \ \ \ &P(3)&\ \ \ \ \ \ \ \ \ &is& \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1^2 + 2^2 + 3^2 &=& \dfrac{3 \cdot 4 \cdot 7}{6} So let \(k\) be a natural number and assume that \(P(k)\) is true. That is, is 2 in the truth set of \(P(n)\)? Principle of Mathematical Induction Solution and Proof. First principle of mathematical induction For each natural number \(n\), \(1 + 4 + 7 + \cdot\cdot\cdot + (3n - 2) = \dfrac{n(3n -1)}{2}.\), We will prove this proposition using mathematical induction. Then to determine the validity of P(n) for every n, use the following principle: Check whether the given statement is true for n = 1. Which of the following sets are inductive sets? Then P(n) is true for all positive integers n. = (n + 1)! One way of proving statements of this form uses the concept of an inductive set introduced in Preview Activity \(\PageIndex{2}\). We should keep in mind that no matter how many examples we try, we cannot prove this proposition with a list of examples because we can never check if 4 divides \((5^n - 1)\) for every natural number \(n\). The basis step is an essential part of a proof by induction. Just because a conjecture is true for many examples does not mean it will be for all cases. Assume that \(T \subseteq \mathbb{N}\) and assume that \(1 \in T\) and that \(T\) is an inductive set. The statement holds for n = 1, and. But in strong induction, the given statement holds true for all the steps from base to the kth step. In Section 4.2, we will learn how to extend this method to statements of the form \((\forall n \in T) (P(n))\), where \(T\) is a certain type of subset of the integers \(\mathbb{Z}\). Hence we can say that by the principle of mathematical induction this statement is valid for all natural numbers n. Show that 22n-1 is divisible by 3 using the principles of mathematical induction. -1 is divisible by 3 using the principles of mathematical induction. So 3 is divisible by 3. The two open sentences in Preview Activity \(\PageIndex{1}\) appeared to be true for all values of \(n\) in the set of natural numbers, \(\mathbb{N}\). &=& \dfrac{k(k + 1)(2k + 1) + 6(k + 1)^2}{6}\\ A class of integers is called hereditary if, whenever any integer x belongs to the class, the successor of x (that is, the integer x + 1) also belongs to the class. Sometimes it helps to look at some specific examples such as \(P(2)\) and \(P(3)\). For each natural number \(n\), we let \(P(n)\) be. Since \(5^{k+1} = 5 \cdot 5^k\), multiply both sides of the congruence \(5^k \equiv 1\) (mod 4) by 5. Now for the general case, if \(k \in \mathbb{N}\), we look at \(P(k + 1)\) and compare it to \(P(k)\). 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