Homework Help is Here – Start Your Trial Now!
Math
Advanced Math
(1 point) By dragging statements from the left column to the right column below, give a proof by induction of the following statement: The correct proof will use 8 of the statements below. Statements to choose from: k(k + 1)(2k + 1) Then 12+22+... +k² = = Thus P(k) is true for all k. Now assume that P (k) is true for an arbitrary Integer k ≥ 1. So 12+22++(k+1)² k(k+1)(2k+1) +(k+1)² k(k+1)(2k+1)+(k+1)² = (k+1)((k+1)+ 1)(2(k + 1) + 1) = Let P(n) be the predicate, 12 +22+ +7 = n(n + 1)(2n+1), 6 Therefore, by the Principle of Mathematical Induction, P(n) is true for all n ≥ 1. Note that 12 KLX1+1)(2-1+1) 6 Thus P(k+1) is true. Then k(k+1)(2k+1) 12+2²++ ++(k+1)²= by the inductive hypothesis. Note that 12+22++(k+1)²= (12+22++k²)+(k+1)². Now assume that P(k + 1) is true for all k. Then we see that 12+2++(k+1)² = k(k+1)(2k+1) +(k+1)² k(k+1)(2k+1) 6(k+1)² + 6 6 2k3 + 3k² + k 6k² + 12k +6 = + 6 6 2k3 +9k2+13k+6 6 (k+1)(k+2)(2k+3) 6 (k+1)((k+1)+ 1)(2(k+1)+1) Your Proof: Put chosen statements in order in this column and press the Submit Answers button. For all n ≥ 1, 12+22+...+n² n(n + 1)(2n+1) = 6

(1 point) By dragging statements from the left column to the right column below, give a proof by induction of the following statement: The correct proof will use 8 of the statements below. Statements to choose from: k(k + 1)(2k + 1) Then 12+22+... +k² = = Thus P(k) is true for all k. Now assume that P (k) is true for an arbitrary Integer k ≥ 1. So 12+22++(k+1)² k(k+1)(2k+1) +(k+1)² k(k+1)(2k+1)+(k+1)² = (k+1)((k+1)+ 1)(2(k + 1) + 1) = Let P(n) be the predicate, 12 +22+ +7 = n(n + 1)(2n+1), 6 Therefore, by the Principle of Mathematical Induction, P(n) is true for all n ≥ 1. Note that 12 KLX1+1)(2-1+1) 6 Thus P(k+1) is true. Then k(k+1)(2k+1) 12+2²++ ++(k+1)²= by the inductive hypothesis. Note that 12+22++(k+1)²= (12+22++k²)+(k+1)². Now assume that P(k + 1) is true for all k. Then we see that 12+2++(k+1)² = k(k+1)(2k+1) +(k+1)² k(k+1)(2k+1) 6(k+1)² + 6 6 2k3 + 3k² + k 6k² + 12k +6 = + 6 6 2k3 +9k2+13k+6 6 (k+1)(k+2)(2k+3) 6 (k+1)((k+1)+ 1)(2(k+1)+1) Your Proof: Put chosen statements in order in this column and press the Submit Answers button. For all n ≥ 1, 12+22+...+n² n(n + 1)(2n+1) = 6

College Algebra (MindTap Course List)
College Algebra (MindTap Course List)
12th Edition
ISBN:9781305652231
Author:R. David Gustafson, Jeff Hughes
Publisher:R. David Gustafson, Jeff Hughes
Chapter8: Sequences, Series, And Probability
Section8.5: Mathematical Induction
Problem 42E
See similar textbooks
icon
Related questions
Question

Don't use chat gpt It Chatgpt means downvote 

(1 point) By dragging statements from the left column to the right column below, give a proof by induction of the following statement: The correct proof will use 8 of the statements below. Statements to choose from: k(k + 1)(2k + 1) Then 12+22+... +k² = = Thus P(k) is true for all k. Now assume that P (k) is true for an arbitrary Integer k ≥ 1. So 12+22++(k+1)² k(k+1)(2k+1) +(k+1)² k(k+1)(2k+1)+(k+1)² = (k+1)((k+1)+ 1)(2(k + 1) + 1) = Let P(n) be the predicate, 12 +22+ +7 = n(n + 1)(2n+1), 6 Therefore, by the Principle of Mathematical Induction, P(n) is true for all n ≥ 1. Note that 12 KLX1+1)(2-1+1) 6 Thus P(k+1) is true. Then k(k+1)(2k+1) 12+2²++ ++(k+1)²= by the inductive hypothesis. Note that 12+22++(k+1)²= (12+22++k²)+(k+1)². Now assume that P(k + 1) is true for all k. Then we see that 12+2++(k+1)² = k(k+1)(2k+1) +(k+1)² k(k+1)(2k+1) 6(k+1)² + 6 6 2k3 + 3k² + k 6k² + 12k +6 = + 6 6 2k3 +9k2+13k+6 6 (k+1)(k+2)(2k+3) 6 (k+1)((k+1)+ 1)(2(k+1)+1) Your Proof: Put chosen statements in order in this column and press the Submit Answers button. For all n ≥ 1, 12+22+...+n² n(n + 1)(2n+1) = 6
Transcribed Image Text:(1 point) By dragging statements from the left column to the right column below, give a proof by induction of the following statement: The correct proof will use 8 of the statements below. Statements to choose from: k(k + 1)(2k + 1) Then 12+22+... +k² = = Thus P(k) is true for all k. Now assume that P (k) is true for an arbitrary Integer k ≥ 1. So 12+22++(k+1)² k(k+1)(2k+1) +(k+1)² k(k+1)(2k+1)+(k+1)² = (k+1)((k+1)+ 1)(2(k + 1) + 1) = Let P(n) be the predicate, 12 +22+ +7 = n(n + 1)(2n+1), 6 Therefore, by the Principle of Mathematical Induction, P(n) is true for all n ≥ 1. Note that 12 KLX1+1)(2-1+1) 6 Thus P(k+1) is true. Then k(k+1)(2k+1) 12+2²++ ++(k+1)²= by the inductive hypothesis. Note that 12+22++(k+1)²= (12+22++k²)+(k+1)². Now assume that P(k + 1) is true for all k. Then we see that 12+2++(k+1)² = k(k+1)(2k+1) +(k+1)² k(k+1)(2k+1) 6(k+1)² + 6 6 2k3 + 3k² + k 6k² + 12k +6 = + 6 6 2k3 +9k2+13k+6 6 (k+1)(k+2)(2k+3) 6 (k+1)((k+1)+ 1)(2(k+1)+1) Your Proof: Put chosen statements in order in this column and press the Submit Answers button. For all n ≥ 1, 12+22+...+n² n(n + 1)(2n+1) = 6
Expert Solution
steps

Step by step

Solved in 2 steps with 2 images

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer
Recommended textbooks for you
College Algebra (MindTap Course List)
College Algebra (MindTap Course List)
Algebra
ISBN:
9781305652231
Author:
R. David Gustafson, Jeff Hughes
Publisher:
Cengage Learning
Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
Algebra
ISBN:
9781133382119
Author:
Swokowski
Publisher:
Cengage
Linear Algebra: A Modern Introduction
Linear Algebra: A Modern Introduction
Algebra
ISBN:
9781285463247
Author:
David Poole
Publisher:
Cengage Learning
College Algebra
College Algebra
Algebra
ISBN:
9781938168383
Author:
Jay Abramson
Publisher:
OpenStax
Elements Of Modern Algebra
Elements Of Modern Algebra
Algebra
ISBN:
9781285463230
Author:
Gilbert, Linda, Jimmie
Publisher:
Cengage Learning,
SEE MORE TEXTBOOKS