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Order 7 of the following sentences so that they form a proof for the following statment: Choose from these sentences: By the induction hypothesis, we have n(n!) < (n + 1)!. For every positive integer n, 1(1!) + 2(2!) + ··· + n(n!) = (n + 1)! - 1. Now we will show that the statement is also true for n + 1. We use induction on n. By the induction hypothesis, Your Proof:

Order 7 of the following sentences so that they form a proof for the following statment: Choose from these sentences: By the induction hypothesis, we have n(n!) < (n + 1)!. For every positive integer n, 1(1!) + 2(2!) + ··· + n(n!) = (n + 1)! - 1. Now we will show that the statement is also true for n + 1. We use induction on n. By the induction hypothesis, Your Proof:

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Order 7 of the following sentences so that they form a proof for the following statment: Choose from these sentences: For every positive integer n, 1(1!) + 2(2!) + ··· · + n(n!) = (n + 1)! — 1. By the induction hypothesis, we have n(n!) < (n + 1)!. Now we will show that the statement is also true for n + 1. We use induction on n. By the induction hypothesis, 1(1!) + 2(2!) + ··· + n(n!) + (n + 1)((n + 1)!) = ((n + 1)! − 1) + (n + 1) ((n + 1)!) = (n + 2)(n + 1)! − 1 = (n + 2)! - 1. Now we consider the case when n is odd. For the inductive step, let n ≥ 1 and suppose that the statement is true for n. Therefore, the statement is also true for n + 1. By induction the statement holds for all positive integers n. First, we consider the case when ʼn is even. For the basis step, if n = 1, then the statement is true because 1(1!) 1 and (1+1)! − 1=1. = Your Proof:
Transcribed Image Text:Order 7 of the following sentences so that they form a proof for the following statment: Choose from these sentences: For every positive integer n, 1(1!) + 2(2!) + ··· · + n(n!) = (n + 1)! — 1. By the induction hypothesis, we have n(n!) < (n + 1)!. Now we will show that the statement is also true for n + 1. We use induction on n. By the induction hypothesis, 1(1!) + 2(2!) + ··· + n(n!) + (n + 1)((n + 1)!) = ((n + 1)! − 1) + (n + 1) ((n + 1)!) = (n + 2)(n + 1)! − 1 = (n + 2)! - 1. Now we consider the case when n is odd. For the inductive step, let n ≥ 1 and suppose that the statement is true for n. Therefore, the statement is also true for n + 1. By induction the statement holds for all positive integers n. First, we consider the case when ʼn is even. For the basis step, if n = 1, then the statement is true because 1(1!) 1 and (1+1)! − 1=1. = Your Proof:
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